In this chapter you will be introduced to the concept of work and the related concepts of kinetic energy and potential energy. After completion of the chapter you will be able to calculate the work done by constant forces, the kinetic energy of a moving object, the gravitational potential energy of an object, and the power developed by a force.
In addition, you will be able to use the principle of conservation of mechanical energy to solve a variety of problems in a much simpler way than you have done previously.
Important Terms
Work The work done by a CONSTANT force acting on an object is the component of the force along the displacement of the object times the magnitude of the displacement.
Kinetic energy
Gravitational potential energy
The energy of an object due to its motion.
The energy of an object due to its position relative to the earth.
Total mechanical energy
The sum of the kinetic and potential energies of an object.
Conservative force A force which does work on an object which is independent of the path taken by the object between its starting point and its ending point.
Average power The work done by a force on an object divided by the time taken to do the work.
The work done on an object by a force, F , is
(6.1) where s is the magnitude of the displacement of the object and θ is the angle between the force and the displacement.
The work done by gravity on an object is
(6.4) where m is the mass of the object, is the initial height of the object, and is the final height of the object.

The work done by nonconservative forces acting on an object is
(6.6)
(6.7a)
(6.7b)
The kinetic energy of an object of mass, m, and speed, v, is
(6.2)
The gravitational potential energy of an object a height, h, above the reference level is
The total mechanical energy of an object is the sum of its kinetic and potential energies.
(6.5)
The work done by nonconservative forces produces a change in the total mechanical energy of the object.
(6.8)
The average power developed by a force which has done work, W, in time, t, is
(6.10a)
The average power developed by a force of magnitude F, moving an object with an average speed, is
(6.11)
Work-Energy Theorem - the total work done by all forces acting on an object is
(6.3)
Principle of Conservation of Mechanical Energy - the total mechanical energy of an object remains constant as the object moves, provided that no net work is done by nonconservative forces.
(6.9a)

The concept of work plays an important role in physics since it connects, via the work-energy theorem, Newton's second law to the important scalar quantities of kinetic energy and potential energy.
Example 1 Positive and negative work
A constant force of 40.0 N is needed to keep a car traveling at constant speed as it moves 5.0 km down the road. How much work is done? Is the work done on or by the car?
The force needed to keep the car moving is in the direction of the car's displacement. Equation
(6.1) gives the work.
Example 2
There must be a retarding force acting on the car in example 1 if it is to remain traveling at constant speed. What is this force? How much work does the force do? Is this work done on or by the car?
Newton's second law applied along the direction of motion of the car gives or
The work done by the retarding force is then given by (6.1) to be
Example 3
A force is pulling an 85.0 kg block across a horizontal surface. The force acts at an angle of 20.0° above the surface. The coefficient of kinetic friction is 0.200, and the block moves a distance of 8.00 m. Find (a) the work done by the pulling force, (b) the work done by the kinetic frictional force, (c) the total work done by all the forces.
(a) The work done by the pulling force is

(b) The magnitude of the frictional force,
Newton's second law to the free body diagram.
, acting on the block is found by applying
In the vertical direction then
The work done by this force is
(c) The total work done by all forces is
Note that the work due to the gravitational force and the normal force is zero since these forces act at a 90° angle to the motion of the block.
Net work done on an object always produces a change in the kinetic energy of the object according to the work-energy theorem (6.3). The theorem is often very useful in finding the speed of an object if the net work done by the forces acting on it are known or can be calculated.
Example 4 Using the work-energy theorem to find the speed of an object
If the block in example 3 starts from rest, what is the speed of the block after it has traveled 8.00 m?

The work-energy theorem gives
Solving for gives
Another common use of the work-energy theorem is finding information about the forces acting on an object when information about the object's speed changes is known.
Example 5 Finding an average force when the speeds are known
A baseball pitcher can throw a 9.00 ounce baseball with speed measured by a radar gun of 90.0 miles per hour (132 ft/s). Assuming that the force exerted by the pitcher on the ball acts over a distance of two arm lengths, each 28.0 inches, what is the average force exerted by the pitcher on the ball?
The work-energy theorem gives if the ball starts at rest. The work done is , so
The gravitational force is one of a class of “special” forces called conservative forces. What makes the gravitational force special is that the work done on an object by gravity depends only on the initial and final positions of the object. The work does NOT depend on how the object got from the initial to the final position.
Example 6 Work done on an object by gravity
How much work is done by gravity when a 5.0 kg object moves 1.5 m horizontally and then moves 3.0 m vertically upward? Vertically downward?
The work done by gravity during the horizontal motion is
The work done by gravity during the vertical motion is
If the object is moving vertically downward, the work done during the vertical motion is

As the previous example shows, the magnitude of the work done by gravity on the object when it is going up is the same as the magnitude of the work done by gravity on the object when it is going down. The negative of the work done by gravity is called the potential energy of the object. The work-energy theorem then gives rise to equation 6.6).
Example 7 Using gravitational potential energy
A 3.00 kg model rocket is launched vertically upward with sufficient initial speed to reach a height of , even though air resistance (a non-conservative force) performs of work on the rocket. How high would the rocket have gone if there were no air resistance?
The equation (6.7b) written for the actual motion of the rocket gives
If the rocket were launched with the same initial kinetic energy and no air resistance acts, then
(6.7b) is or
If no friction or other non-conservative force acts on an object then its total mechanical energy remains constant. This principle is very useful in solving problems involving the motion of an object. It is often easier to use than Newton's laws or the kinematic equations since it involves scalars rather than vectors. Also, it only requires that you have knowledge about the motion at the beginning and end.
Example 8
A truck is descending a winding mountain road. When the truck is 1480 m above sea level and traveling 15 m/s, its brakes fail. What is the maximum possible speed of the truck at the foot of the mountain 550 m above sea level?

The truck would have the maximum possible speed if friction were ignored. In this case, the total mechanical energy of the truck is conserved as it moves down the mountain road.
Taking the zero of potential energy at sea level we have
The average power is the rate at which work is done as given by equation (6.10a).
Example 9
A 73 kg sprinter, starting from rest, reaches a speed of 7.0 m/s in 1.8 s with negligible effect due to air resistance. The sprinter then runs the remainder of the race at a steady speed of 7.0 m/s under the influence of a 35 N force due to air resistance. Calculate the average power needed (a) to accelerate the runner and (b) to sustain the steady speed at which most of the race is run.
(a) In order to find the power, we need to find the work needed to accelerate the runner. The work-energy theorem gives so
(b) The force that the sprinter must exert to run at constant speed is equal in magnitude and opposite in direction to the force exerted on him by air resistance. The power needed to run the race at a constant speed is then

Practice Problems
1 The driver of a 1500-kg car slams on the brakes locking the wheels. A total retarding force of 1800 N acts to stop the car in a distance of 70.0 m. How much work is done in bringing the car to a halt? Is the work done on or by the car?
2 A 4200-kg truck descending a 5.0° hill is brought to a stop in 95 m. The driver applies the brakes so that the wheels lock. (a) If the coefficient of kinetic friction between the truck tires and the road is 0.60, how much work is done by friction in stopping the truck? (b) How much work is done on the truck by gravity?
3 In problem 1, how fast was the car traveling initially?
4 In problem 2, how fast was the truck traveling immediately before the brakes were applied?
5 A bicyclist tops the crest of a 20.0 m high hill moving with a speed of 1.0 m/s.
Ignoring friction find the speed of the bicycle at the bottom of the hill.
6 A 0.60 kg ball thrown vertically upward with an initial speed of 12.0 m/s rises to a height of 5.1 m. What is the average force exerted by the air on the ball? How high would the ball have gone if no air resistance acted on it?
7 A swing is observed to rise to a maximum height of 1.5 m above its lowest point.
How fast is it going at its lowest point? Ignore friction and air resistance.
8 A 1500 kg car accelerates from rest to 75 km/h in 45 s. How much power is supplied by the engine to accelerate the car?