Wednesday, June 16, 2021

We use the term ‘work’ in everyday conversation to mean many different things. We talk about going to work, doing homework, working in class. Physicists mean something very specific when they talk about work. In Physics we use the term work to describe the process of transferring energy from object or system to another or converting energy from one form to another. You will learn that work and energy are closely related to Newton’s laws of motion. You shall see that the energy of an object is its capacity to do work and doing work is the process of transferring energy from one object or form to another by means of a force. In other words,

  • an object with lots of energy can do lots of work.
  • when object A transfers energy to object B, the energy of object A decreases by the same amount as the energy of object B increases, we say that object A does work on object B.

Lifting objects or throwing them requires that you do work on them. Even making an electrical current flow requires that something do work. Objects or systems must have energy to be able to do work on other objects or systems by transferring some of their energy. 
Key linked concepts

  • Units and unit conversions — Physical Sciences, Grade 10, Science skills
  • Equations — Mathematics, Grade 10, Equations and inequalities
  • Techniques of vector addition —Physical Sciences, Grade 10, Vectors and scalars
  • Newton’s laws — Physical Sciences, Grade 11, Forces
  • Force diagrams — Physical Sciences, Grade 11, Forces

We cover different topics in different chapters in different grades but that doesn’t mean that they are not related. In fact, it is very important to note that all of the different topics related to mechanics (forces, mechanical energy, momentum, rectilinear motion) actually form a consistent picture of the same physical system. There have been
examples where we’ve shown the same results using two methods, for example determining speed or velocity using equations of motion or conservation of mechanical energy. Learning about work will help us tie everything we’ve learnt about previously together. 
Work will allow us to connect energy transfer to forces, which we have already linked to momentum and the equations of motion. When a force tends to act in or against the direction of motion of an object we say that the force is doing work on the object. Specifically, work is defined mathematically in terms of the force and the displacement of the object.
DEFINITION: Work
When a force acts in or against the direction of motion of an object, work is done on
the object.W = FΔx cos θ

 
 
 
 
 
 

 
 

Important: cos θ tells you the relative direciton of the force and the displacment which is important. If the component of the force along the direction of the displacement is opposite in direction to the displacement then the sign of the displacement vector and force vector will be different. This is regardless of which direction was chosen as a positive direction.
Let us look at some examples to understand this properly. In the images below the grey dot represents an object. A force, ​\( \vec{F} \), acts on the object. The object moves through a displacement, Δ\( \vec{x} \). What is the sign of the work done in each case?


It is only the direction of the force on the object that matters and not the direction from the source of the force to the object. In Figure 5.3 both powerlifters are exerting an upwards force on the weights. On the left the weight is being pulled upwards and on the right it is being pushed upwards.

Weight lifting is a good context in which to think about work because it helps to identify misconceptions introduced by everyday use of the term ‘work’. In the two cases in Figure 5.3 everyone would describe moving the weights upwards as very hardwork. From a physics perspective, as the powerlifters lift the weight they are exerting
a force in the direction of the displacement so positive work is done on the weights.

Consider the strongman walking in Figure 5.4. He carries two very heavy sleds as far as he can in a competition. What work is the man doing on the sleds and why?
Most people would say he is working very hard because the sleds are heavy to carry but from a physics perspective he is doing no work on the sleds. The reason that he does no work is because the force he exerts is directly upwards to balance the force of gravity and the displacement is in the horizontal direction.
Therefore there is no component of the force in the direction of displacement (θ = 90°) and no work done. His muscles do need to use their energy reserves to maintain the force to balance gravity. That does not result in energy transfer to the sleds.
Investigation: Is work done?
Decide whether or not work is done in the following situations. Remember that for work to be done a force must be applied in the direction of motion and there must be a displacement. Identify which two objects are interacting, what the action-reaction pairs of forces are and why the force described is or isn’t doing work.
1. Max pushes against a wall and becomes tired.
2. A book falls off a table and free falls to the ground.
3. A rocket accelerates through space.
4. A waiter holds a tray full of plates above his head with one arm and carries it straight across the room at constant speed.

For each of the above pictures, the force vector is acting in the same direction as the displacement vector. As a result, the angle θ = 0°because there is no difference in angle between the direction of applied force and the direction of displacement.
The work done by a force can then be positive or negative. This sign tells us about the direction of the energy transfer. Work is a scalar so the sign should not be misinterpreted to mean that work is a vector. Work is defined as energy transfer, energy is a scalar quantity and the sign indicates whether energy was increased or decreased.

  • If \( \vec{F} \)applied  acts or has a component acting in the same direction as the motion, then positive work is being done. In this case the object on which the force is applied gains energy. 
  • If the direction of motion and \( \vec{F} \)applied are opposite, then negative work is being done. This means that energy is lost and the object exerting the force gains energy. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you! applied

Worked example 1: Calculating work on a car when speeding up.
QUESTION
A car is travelling along a straight horizontal road. A force of 500 N is applied to the car in the direction that it is travelling, speeding it up. While it is speeding up is covers a distance of 20 m. Calculate the work done on the car.
 
SOLUTION
Step 1: Analyse the question to determine what information is provided

  • The magnitude of the force applied is F = 500 N.
  • The distance moved is Δx = 20 m.
  • The applied force and distance moved are in the same direction. Therefore, the angle between the force and displacement is θ = 0°

These quantities are all in SI units, so no unit conversions are required.
Step 2: Analyse the question to determine what is being asked

  • We are asked to find the work done on the car. We know from the definition that work done is W = FΔx cosθ.

Step 3: Next we substitute the values and calculate the work done
W = FΔx cosθ.
=(500) (20) (cos 0)
=(500) (20) (1)
= 10 000 J
Remember that the answer must be positive, as the applied force and the displacement are in the same direction.
In this case, the car gains kinetic energy.
Worked example 2: Calculating work on the car while braking

 
 
 
 
 
 
 
 
 
 
 
What happens when the applied force and the motion are not parallel? By using the formula W = FΔx cosθ , we are actually calculating the component of the applied force in the direction of motion. Note that the component of the force perpendicular to the direction of motion does no work.
Worked example 3: Calculating work done on a box pulled at an angle

QUESTION
Calculate the work done on a box, if it is pulled 5 m along the ground by applying a force of F = 20 N at an angle of 60° to the horizontal.
SOLUTION
Step 1: Analyse the question to determine what information is provided

  • The force applied is F=20 N
  • The distance moved is Δx = 5 m along the ground
  • The angle between the applied force and the motion is θ=60

These quantities are in the correct units so we do not need to perform any unit conversions.
Step 2: Analyse the question to determine what is being asked
We are asked to find the work done on the box.
Step 3: Substitute and calculate the work done
Now we can calculate the work done on the box:
W = FΔx cos θ
=(20) (5) (cos 60)
= 50 J
Note that the answer is positive as the component of the force parallel to the direction of motion is in the same direction as the motion. The work done on the box is 50J.
Exercise 5 – 1: Work

1.A 10 N force is applied to push a block across a frictionless surface for a displacement of 5,0 m to the right. The block has a weight of 20 N. Determine the work done by the following forces: normal force, weight , applied force.
2. A 10 N frictional force slows a moving block to a stop after a displacement of 5,0 m to the right. The block has a weight of 20 N Determine the work done by the following forces: normal
force, weight, frictional force.
3. A 10 N force is applied to push a block across a frictional surface at constant speed for a displacement of 5,0 m to the right. The block has a weight of 20 N and the frictional force is 10 N. Determine the work done by the following forces: normal force, weight, applied force and frictional force.
4.An object with a weight of 20 N is sliding at constant speed across a frictionless surface for a displacement of 5 m
to the right. Determine if there is any work done.

5.An object with a weight of 20 N is pulled upward at constant speed by a 20 N force for a vertical displacement of 5m . Determine if there is any work done.
6. Before beginning its descent, a roller coaster is always pulled up the first hill to high initial height. Work is done on the roller coaster to achieve this initial height. A coaster designer is considering three different incline angles of the hill at which to drag the 2000 kg car train to the top of the 60 m high hill. In each case, the force applied to the car will be applied parallel to the hill. Her critical question is: which angle would require the least work? Analyse the data, determine the work done in each case, and answer this critical question.

7. A traveller carries a 150 N suitcase up four flights of stairs (a total height of 12 m) and then pushes it with a horizontal force of 60 N at a constant speed of 0,25 m.s-1 for a horizontal distance of 50 m on a frictionless surface. How much work does the traveller do on the suitcase during this entire trip?

8. A parent pushes down on a pram with a force of 50 N at an angle of 30 to the horizontal. The pram is moving on a frictionless surface. If the parent pushes the pram for a horizontal distance of 30 m, how work is done on the pram?
9. How much work is done by the force required to raise a 2000 N lift 5 floors vertically at a constant speed? The vertical distance between floors is 5 m high.
10. A student with a mass of 60 kg runs up three flights of stairs in 15 s, covering a vertical distance of 10 m. Determine the amount of work done by the student to elevate her body to this height.
Net work
We have only looked at a single force acting on an object. Sometimes more than one force acts at the same time (we dealt with this in Grade 11). We call the work done after taking all the forces into account the net work done. If there is only one force acting then the work it does, if any, is the net work done. In this case there are two equivalent approaches we can adopt to finding the net work done on the object. We can:

  • Approach 1: calculate the work done by each force individually and then sum them taking the signs into account. If one force does positive work and another does the same amount of work but it is negative then they cancel out.
  • Approach 2: calculate the resultant force from all the forces acting and calculate the work done using the resultant force. This will be equivalent to Approach 1.
    If the resultant force parallel to the direction of motion is zero, no net work will be done.

Remember that work done tells you about the energy transfer to or from an object by means of a force. That is why we can have zero net work done even if multiple large forces are acting on an object. Forces that result in positive work increase the energy of the object, forces that result in negative work reduce the energy of an object. If as
much energy is transferred to an object as is transferred away then the final result is that the object gains no energy overall.
Worked example 4: Approach 1, calculating the net work on a car
QUESTION

The same car is now accelerating forward, but friction is working against the motion of the car. A force of 300 N is applied forward on the car while it is travelling 20 m forward. A frictional force of 100 N acts to oppose the motion. Calculate the net work done on the car.
Only forces with a component in the plane of motion are shown on the diagram. No work is done by Fg or FNormal as they act perpendicular to the direction of motion.
SOLUTION
Step 1: Analyse the question to determine what information is provided

  • The force applied is Fapplied=300 N forwards.
  • The force of friction is Ffriction=100 N opposite to the direction of motion.
  • The distance moved is Δx = 20 m.

The applied force and distance moved are in the same plane so we can calculate friction the work done by the applied forward force and the work done by the force of friction backwards.
Step 2: Analyse the question to determine what is being asked
We are asked to find the net work done on the car. We know from the definition that work done is W = FΔx cosθ 

As mentioned before, there is an alternative method to solving the same problem, which is to determine the net force acting on the car and to use this to calculate the work. This means that the vector forces acting in the plane of motion must be added to get the net force ​\( \vec{F} \)net. The net force is then applied over the displacement to get the net work Wnet. 
Worked example 5: Approach 2, calculating the net force
QUESTION
The same car is now accelerating forward, but friction is working against the motion of the car. A force of 300 N is applied forward on the car while it is travelling 20m forward. A frictional force of 100 N acts to oppose the motion. Calculate the net work
done on the car.
SOLUTION
Step 1: Analyse the question to determine what information is provided

  • The force applied is ​\( \vec{F} \)applied=300 N forwards.
  • The force of friction is  ​\( \vec{F} \)friction=100 N backwards.
  • The distance moved is Δx = 20 m.
  • The applied forces ​\( \vec{F} \)applied = 300 N and the force of friction ​\( \vec{F} \)friction= 100 N are in the same plane as the distance moved. Therefore, we can add the vectors. As vectors require direction, we will say that forward is positive and therefore backward is negative. Note, the force of friction is acting at 180° friction i.e. backwards and so is acting in the opposite vector direction i.e. negative.

These quantities are all in the correct units, so no unit conversions are required.
Step 2: Analyse the question to determine what is being asked

  • We are asked to find the net work done on the car. We know from the definition that work done is Wnet=FnetΔxcosθ

Step 3: We calculate the net force acting on the car, and we convert this into net work.
First we draw the force diagram:
Let forwards (to the left in the picture) be positive. We know that the motion of the car is in the horizontal direction so we can neglect the force due to gravity, ​\( \vec{F} \)g , and the normal force, ​\( \vec{N} \)​. Note: if the car were on a slope we would need to calculate the component of gravity parallel to the slope. 
\( \vec{F} \)net=​\( \vec{F} \)applied+​\( \vec{F} \)friction
=(+300)+(-100) 
\( \vec{F} \)​ =​ 200 N forwards
 
\( \vec{F} \)net is pointing in the same direction as the displacement, therefore the angle between the force and displacement is θ = 0. 
IMPORTANT!
The two different approaches give the same result but it is very important to treat the signs correctly. The forces are vectors but work is a scalar so they shouldn’t be interpreted in the same way.
Wnet= FnetΔxcosθ
=(200) (20)cos(0)
= 4000 J
 
 

Conservative and non-conservative forces
In Grade 10, you saw that mechanical energy was conserved in the absence of non-conservative forces. It is important to know whether a force is an conservative force or an non-conservative force in the system, because this is related to whether the force can change an object’s total mechanical energy when it does work on an object.
When the only forces doing work are conservative forces (for example, gravitational and spring forces), energy changes forms – from kinetic to potential (or vice versa); yet the total amount of mechanical energy (EK+ Ep) is conserved. For example, as an object falls in a gravitational field from a high elevation to a lower elevation, some of the object’s potential energy is changed into kinetic energy. However, the sum of the kinetic and potential energies remain constant.
Investigation: Non-conservative forces
We can investigate the effect of non-conservative forces on an object’s total mechanical energy by rolling a ball along the floor from point A to point B.

In the absence of friction and other non-conservative forces, the ball should slide along the floor and its speed should be the same at positions A and B. Since there are no non-conservative forces acting on the ball, its total mechanical energy at points A and B are equal.

Now, let’s investigate what happens when there is friction (an non-conservative force) acting on the ball.
Roll the ball along a rough surface or a carpeted floor. What happens to the speed of the ball at point A compared to point B? 
If the surface you are rolling the ball along is very rough and provides a large non-conservative frictional force, then the ball should be moving much slower at point B than at point A. 
Let’s compare the total mechanical energy of the ball at points A and B:

However, in this case, VA ≠ VB and therefore ETotal,A ≠ ETotal,B. Since

VA>VB

ETotal,A> ETotal,B

Therefore, the ball has lost mechanical energy as it moves across the carpet. However,although the ball has lost mechanical energy, energy in the larger system has still been conserved. In this case, the missing energy is the work done by the carpet through applying a frictional force on the ball. In this case the carpet is doing negative work on
the ball.

When an non-conservative force (for example friction, air resistance, applied force) does work on an object, the total mechanical energy (EK+ Ep)  of that object changes. If positive work is done, then the object will gain energy. If negative work is done, then the object will lose energy.
When a net force does work on an object, then there is always a change in the kinetic energy of the object. This is because the object experiences an acceleration and therefore a change in velocity. This leads us to the work-energy theorem.
DEFINITION: Work-energy theorem
The work-energy theorem states that the work done on an object by the net force is equal to the change in its kinetic energy:                                                     
                                                                                  WNet=ΔEk=Ek,f-Ek,i
Worked example 6: Work-energy theorem
QUESTION
A 1 kg brick is dropped from a height of 10 m. Calculate the work that has been done on the brick between the moment it is released and the moment when it hits the ground. Assume that air resistance can be neglected.
SOLUTION
Step 1: Determine what is given and what is required

  • Mass of the brick: m = 1 kg.
  • Initial height of the brick: hi= 10 m.
  • Final height of the brick: hfi= 0 m.
  • We are required to determine the work done on the brick as it hits the ground.

Step 2: Determine how to approach the problem
The brick is falling freely, so energy is conserved. We know that the work done is equal
to the difference in kinetic energy. The brick has no kinetic energy at the moment it
is dropped, because it is stationary. When the brick hits the ground, all the brick’s
potential energy is converted to kinetic energy.
Step 3: Determine the brick’s potential energy at hi

Ep= m.g.hi

              =(1) (9,8) (10)

= 98 J

Step 4: Determine the work done on the brick
The brick had 98 J of potential energy when it was released and 0 J of kinetic energy. When the brick hit the ground, it had 0 J of potential energy and 98 J of kinetic energy.
Therefore Ek,i= 0 J and Ek,f= 98 J.
From the work-energy theorem:

Wnet = ΔEk

          = Ek,fEk,i

          = 98 – 0

           = 98 J

Hence, 98 J of work was done on the brick.
Worked example 7: Work-energy theorem 2
QUESTION
The driver of a 1000 kg car travelling at a speed of 16,7 m.s-1 applies the car’s brakes when he sees a red light. The car’s brakes provide a frictional force of 8000 N. Determine the stopping distance of the car.
SOLUTION
Step 1: Determine what is given and what is required
We are given:

  • mass of the car: m = 1000 kg
  • speed of the car: v = 16,7 m.s-1
  • frictional force of brakes: ​\( \tilde{F} \)= 8000 N

We are required to determine the stopping distance of the car.
Step 2: Determine how to approach the problem
We apply the work-energy theorem. We know that all the car’s kinetic energy is lost to friction. Therefore, the change in the car’s kinetic energy is equal to the work done by the frictional force of the car’s brakes.Therefore, we first need to determine the car’s kinetic energy at the moment of braking using: Ek=½mv2
This energy is equal to the work done by the brakes. We have the force applied by the brakes, and we can use:      W = FΔx cosθ  to determine the stopping distance.

Worked example 8: Block on an inclined plane [credit: OpenStax College]
QUESTION
A block of 2 kg is pulled up along a smooth incline of length 10 m and height 5 m by applying an non-conservative force. At the end of incline, the block is released from rest to slide down to the bottom. Find the

  1. work done by the non-conservative force,
  2. the kinetic energy of the block at the end of round trip, and
  3. the speed at the end of the round trip.


We have represented the non-conservative force on the force diagram with an arbitrary vector\( \tilde{F} \)​ acts only during upward journey. Note that the block is simply released at the end of the upward journey. We need to find the work done by the non-conservative force only during the upward journey.
                        WF=WF(up)+WF(down)=WF(up)+0=WF(up)
The kinetic energies in the beginning and at the end of the motion up the slope are zero.
We can conclude that sum of the work done by all three forces is equal to zero during the upward motion. The change in kinetic energy is zero which implies that the net work done is zero.
                                           WNet = WF(up)+Wg(up)+WN(up)
                                                         0 = WF(up)+Wg(up)+WN(up)
If we know the work done by the other two forces (normal force and gravity), then we can calculate the work done by the non-conservative force, F, as required.
Step 3: Work done by normal force during upward motion
The block moves up the slope, the normal force is perpendicular to the slope and,
therefore, perpendicular to the direction of motion. Forces that are perpendicular to the direction of motion do no work. 
                                                    WNet = WF(up)+Wg(up)+WN(up)
                                                         0 = WF(up)+Wg(up)+WN(up)
                                                     WF(up) = -Wg(up)

IMPORTANT!


Be careful not to be confused by which angle has been labelled α and which θ. α is not the angle between the force and the direction of motion but the incline of the plane in this particular problem. It is important to understand which symbol represents which physical quantity in the equations you have learnt.
Hence, the work done by the non-conservative force during the round trip is 

        WF(up) =  WF(up)= – Wg(up)

= – (-98)

= 98 J

Step 5: Kinetic energy at the end of round trip
The kinetic energy at the end of the upward motion was zero but it is not zero at the end of the entire downward motion.
We can use the work-energy theorem to analyse the whole motion:

        W(round trip) =  Ek,f– Ek,i

= Ek,f– 0

= Ek,f


Exercise 5 – 2: Energy
1. Fill in the table with the missing information using the positions of the 1 kg ball in the diagram below combined with the work-energy theorem.


2. A falling ball hits the ground at 10 m.s-1 in a vacuum. Would the speed of the ball be increased or decreased if air resistance were taken into account. Discuss using the work-energy theorem.
3. A pendulum with mass 300 g is attached to the ceiling. It is pulled up to point A which is a height h = 30 cm from the equilibrium position.

Calculate the speed of the pendulum when it reaches point B (the equilibrium point). Assume that there are no non-conservative forces acting on the pendulum.

There are two categories of forces we will consider, conservative and non-conservative.
DEFINITION: Conservative force
A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken.
A conservative force results in stored or potential energy and we can define a potential energy (Ep ) for any conservative force.
Gravity is a conservative force and we studied gravitational potential energy in Grade
10. We now have all the concepts we need to actually deduce this ourselves. Let us consider pushing a ball up a number of different slopes.

Figure 5.5: Three different slopes are shown, all rising to a height of h . The imaginary right- angled triangle is shown for each slope. d is the length of the slope. α is the angle the slope makes with the horizontal.
The slope, of length d is the hypotenuse of an imaginary right-angled triangle. The work done by gravity while pushing a ball of mass, m , up each of the slopes can be

This final result is independent of the angle of the slope. This is because sinα  opposite/hypotenuse=h/d and so the distance cancels out. If the ball moves down the slope the only change is the sign, the work done by gravity still only depends on the change in height. This is why mechanical energy includes gravitational potential energy and is conserved. If an object goes up a distance h gravity does negative work, if it moves back down h gravity does positive work, but the absolute amount of work is the same so you ‘get it back’, no matter what path you take!
This means that the work done by gravity will be same for the ball moving up any
of the slopes because the end position is at the same height. The different slopes do
not end in exactly the same position in the picture. If we break each slope into two
sections as show in Figure 5.6 then we have 3 different paths to precisely the same
end-point. In this case the total work done by gravity along each path is the sum of the
work done on each piece which is just related to the height. The total work done is
related to the total height.

Figure 5.6: Three different paths that lead from the same start-point to the same-end point. Each path leads to the same overall change in height, h, and, therefore, the same work done by gravity.
There are other examples, when you wind up a toy, an egg timer, or an old-fashioned watch, you do work against its spring and store energy in it. (We treat these springs as ideal, in that we assume there is no friction and no production of thermal energy.) This stored energy is recoverable as work, and it is useful to think of it as potential energy contained in the spring. The total work done by a conservative force results in a change in potential energy, ΔEp. If the conservative force does positive work then the change in potential energy is negative. Therefore: 

DEFINITION: Non-conservative force
A non-conservative force is one for which work done on the object depends on the path taken by the object.
IMPORTANT!
Non-conservative forces do not imply that total energy is not conserved. Total energy is always conserved. Non-conservative forces mean that mechanical energy isn’t conserved in a particular system which implies that the energy has been transferred in a process that isn’t reversible.
Friction is a good example of a non-conservative force because if removes energy from the system so the amount of mechanical energy is not conserved. Non-conservative forces can also do positive work thereby increasing the total mechanical energy of the system.
The energy transferred to overcome friction depends on the distance covered and is converted to thermal energy which can’t be recovered by the system.


Non-conservative forces and work-energy theorem
We know that the net work done will be the sum of the work done by all of the individual forces:

When the non-conservative forces oppose the motion, the work done by the non-conservative forces is negative, causing a decrease in the mechanical energy of the system. When the non-conservative forces do positive work, energy is added to the system. If the sum of the non-conservative forces is zero then mechanical energy is conserved.
Worked example 9: Sliding footballer [credit: OpenStax College Physics]

 
QUESTION
Consider the situation shown where a
football player slides to a stop on level ground. Using energy considerations,
calculate the distance the 65,0 kg football
player slides, given that his initial
speed is 6,00 m.s-1 and the force
of friction against him is a constant
450 N.
SOLUTION
     Step 1: Analyse the problem and determine what is given
Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because F is in the opposite direction of the motion (that is,θ = 180 of, and so cos θ = 1). Thus Wnon-conservative = –FfΔx
There is no change in potential energy.

Step 3: Quote the final answer
The footballer comes to a stop after sliding for 2,60 m.
Discussion
The most important point of this example is that the amount of non conservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.
Worked example 10: Sliding up a slope [credit: OpenStax College Physics]
QUESTION
The same 65,0 kg footballer running at the same speed of 6,00 m.s-1 dives up the
inclined embankment at the side of the field. The force of friction is still 450 N as it is
the same surface, but the surface is inclined at 5o . How far does he slide now?
SOLUTION
Step 1: Analyse the question
Friction stops the player by converting his kinetic energy into other forms, including
thermal energy, just in the previous worked example. The difference in this case is that
the height of the player will change which means a non-zero change to gravitational
potential energy.
The work done by friction is negative, because Ff is in the opposite direction of the motion (that is, θ = 180).
We sketch the situation showing that the footballer slides a distance d up the slope.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
As might have been expected, the footballer slides a shorter distance by sliding uphill.Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance d that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy instead, we need only consider the gravitational potential energy, without combining and resolving force vectors. This simplifies the solution considerably.
Exercise 5 – 3: Energy conservation
1. A 60,0 kg skier with an initial speed of 12,0 m.s-1
coasts up a 2,50 m-high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0,0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

2. a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km.h-1 ?
b) If, in actuality, a 750 kg car with an initial speed of 110 km.h-1 is observed to coast up a hill to a height 22,0 m above its starting point, how much thermal energy was generated by friction?
c) What is the average force of friction if the hill has a slope 2,5 above the horizontal?
3. A bullet traveling at 100 m/s just pierces a wooden plank of 5 m. What should be the speed (in m/s) of the bullet to pierce a wooden plank of same material, but having a thickness of 10m?
 

Now that we understand the relationship between work and energy, we are ready to look at a quantity related the rate of energy transfer. For example, a mother pushing a trolley full of groceries can take 30 s or 60 s to push the trolley down an aisle. She does the same amount of work, but takes a different length of time. We use the idea of power to describe the rate at which work is done
Power is defined as the rate at which work is done or the rate at which energy is transfered to or from a system. The mathematical definition for power is: P = W/t

IMPORTANT!
In the case where the force and the velocity are in opposite directions the power will be negative.
The unit of power is watt (symbol W).
Worked example 11: Power calculation 1
QUESTION
Calculate the power required for a force of 10 N applied to move a 10 kg box at a speed of 1 m·s−1 over a frictionless surface.
SOLUTION
Step 1: Determine what is given and what is required.
• We are given the force, F = 10 N.
• We are given the speed, v = 1 m · s−1 .
• We are required to calculate the power required
Step 2: Draw a force diagram

Step  3:  Determine how  to  approach the problem from the force diagram, we see that the weight of the box is acting at right angles to the  direction  of motion.    The weight does not contribute  to the work done and does not contribute to the power calculation.   We can therefore calculate power from: P = F · v .
Step 4: Calculate the power required
P = F · v
= (10 N) 1 m · s−1
= 10 W
Step 5: Write the final answer
10 W of power are required for a force of 10 N to move a 10 kg box at a speed of 1 m·s−1
Machines are designed and built to do work on objects. All machines usually have a power rating. The power rating indicates the rate at which that machine can do work upon other objects.
A car engine is an example of a machine which is given a power rating. The power rating relates to how rapidly the car can accelerate. Suppose that a 50 kW engine could accelerate the car from 0 km·hr−1 to 60 km·hr−1 in 16 s. Then a car with four times the power rating (i.e. 200 kW) could do the same amount of work in a quarter of the time. That is, a 200 kW engine could accelerate the same car from 0 km·hr−1 to 60 km·hr−1 in 4s.
Worked example 12: Power calculation 2
QUESTION
A forklift lifts a crate of mass 100 kg at a constant velocity to a height of 8 m over a time of 4 s. The forklift then holds the crate in place for 20 s. Calculate how much power the forklift exerts in lifting the crate? How much power does the forklift exert in holding the crate in place?
SOLUTION
Step 1: Determine what is given and what is required
We are given:
• mass of crate: m=100 kg
• height that crate is raised: h=8 m
• time taken to raise crate: tr = 4 s
• time that crate is held in place: ts = 20 sWe are required to calculate the power exerted.
Step 2: Determine how to approach the problem
We can use:

to calculate power. The force required to raise the crate is equal to the weight of the crate.
Step 3: Calculate the power required to raise the crate

Step 4: Calculate the power required to hold the crate in place
While the crate is being held in place, there is no displacement. This means there is no work done on the crate and therefore there is no power exerted.
Step 5: Write the final answer
1960 W of power is exerted to raise the crate and no power is exerted to hold the crate in place.
Worked example 13: Stair climb
QUESTION

Step 3: Quote the final answer
The power generated is 538,0 W.
The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating.
Worked example 14: A borehole
QUESTION
What is the power required to pump water from a borehole that has a depth h =
15,0 m at a rate of 20,0 l·s−1 ?
SOLUTION
Step 1: Analyse the question
We know that we will have to do work on the water to overcome gravity to raise it a certain height. If we ignore any inefficiencies we can calculate the work, and power, required to raise the mass of water the appropriate height.
We know how much water is required in a single second. We can first determine the mass of water: 20,0 l × 1 kg/1l = 20,0 kg.
The water will also have non-zero kinetic energy when it gets to the surface because it needs to be flowing. The pump needs to move 20,0 kg from the depth of the borehole
every second, we know the depth so we know the speed that the water needs to be moving is v = h/t = 15,0/1  = 15,0 m·s−1 .

Experiment: Simple measurements of human power
You can perform various physical activities, for example lifting measured weights or climbing a flight of stairs to estimate your output power, using a stop watch. Note: the human body is not very efficient in these activities, so your actual power will be much greater than estimated here.
Exercise 5 – 4: Power
1. [IEB 2005/11 HG] Which of the following is equivalent to the SI unit of power:
a) V·A
b) V·A−1
c) kg·m·s−1
d) kg·m·s−2
2. Two students, Bill and Bob, are in the weight lifting room of their local gym. Bill lifts the 50 kg barbell over his head 10 times in one minute while Bob lifts the
50 kg barbell over his head 10 times in 10 seconds. Who does the most work? Who delivers the most power? Explain your answers.
3. Jack and Jill ran up the hill. Jack is twice as massive as Jill; yet Jill ascended the same distance in half the time. Who did the most work? Who delivered the most power? Explain your answers.
4. When doing a chin-up, a physics student lifts her 40 kg body a distance of 0,25 m in 2 s. What is the power delivered by the student’s biceps?
5. The unit of power that is used on a monthly electricity account is kilowatt-hours (symbol kWh). This is a unit of energy delivered by the flow of 1 kW of electricity for 1 hour. Show how many joules of energy you get when you buy 1 kWh of electricity.
6. An escalator is used to move 20 passengers every minute from the first floor of a shopping mall to the second. The second floor is located 5-meters above the first floor. The average passenger’s mass is 70 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.



Exercise 5  –  5:

  1. How much work does a person do in pushing a shopping trolley with a force of
    200 N over a distance of 80 m in the direction of the force?
  2. How much work does the force of gravity do in pulling a 20 kg box down a 45◦
    frictionless inclined plane of length 18 m?
  3. [IEB 2001/11 HG1] Of which one of the following quantities is kg·m2 ·s−3 the base S.I. unit?


4. [IEB 2003/11  HG1] A motor is used to raise a mass m through a vertical height h in time t. What is the power of the motor while doing this?

5. [IEB 2002/11 HG1] An electric motor lifts a load of mass M vertically through a height h at a constant speed v. Which of the following expressions can be used to calculate the power transferred by the motor to the load while it is lifted at constant speed?

6. A set 193 kg containers need to be lifted onto higher floors during a building operation. The distance that they need to be raised is 7.5 m, at constant speed. The project manager has determined that in order to keep to budget and time this has to happen in as close to 5,0 s as possible. The power ratings for three motors are listed as 1,0 kW, 3,5 kW, and 5,5 kW. Which motor is best for the job?