Work is a scalar quantity measured in Joules. Kinetic energy is a scalar quantity measured in Joules. They are the same kind of number measured with the same units. They are simply two different ways of describing the same thing.

The Work-Kinetic Energy Theorem states that the net work on an object is equal to the change in that object's kinetic energy. That means if you add up the total work done by all the forces on an object, you will know the object's change in KE.

W_net = ΔKE

Let's take a closer look at both sides of this equation. On the left side we have Wnet. Wnet means the total work done on a system. Either add up each individual work done by each individual force, or find the total force and calculate the work done by it. On the right side we have ΔKE. Remember that the Greek symbol Δ simply means change, ΔKE = KE_f – KE_i.

Watch the following animation comparing the stopping distances of three cars moving at different velocities.

Source: Energy, The Physics Classroom

The four values being shown to the right of each car are Kinetic Energy (red), Potential Energy (always zero), Work (yellow), and Total Mechanical Energy (green)

What is friction doing to the velocity of the car in terms of work and energy?

Answer Friction is doing negative work to remove kinetic energy from the car and reduce its velocity. Close

Remember that the work energy theorem predicts the stopping distance by:

Fdcosθ = 1 2 mv_2² – 1 2 mv_1²

We can substitute in that q = 180 since friction is directly opposite displacement and since cos(180) = -1, that makes the left side of the equation:

-Fd

We can also plug in that v2 = 0 because the cars are stopping. That makes the right side of the equation:

- 1 2 mv_1²

The negative signs cancel to yield:

Fd = 1 2 mv_1²

or

d = mv_1² 2F Close

 

Example 1

Calculate the amount of net work required for a 1125 kg car to go from 2.50 m sec to 3.50 m sec .

Disregard air resistance and friction.
net work = ?     mass = 1125 kg     v_i = 2.500 m sec     v_f = 3.500 m sec

The basic equation to be used is W_net = ΔKE.

W_net = ΔKE

Net work equals the change in kinetic energy. So net work equals the final kinetic energy minus the initial kinetic energy. Thus net work equals the quantity one-half times the mass times the final velocity squared minus the quantity one-half times the mass times the initial velocity squared. Using the numbers from our problem, net work equals the quantity one-half times 1125 kg times the square of 3.500 meters per second minus the quantity one-half times 1125 kg times the square of 2.500 meters per second. Therefore, net work equals 3375 Joules.

= KE_f – KE_i
= 1 2 mv_f² – 1 2 mv_i²
=( 1 2 )(1125 kg)(3500 m sec )² – ( 1 2 )(1125 kg)(2.500 m sec )²
= 3375 J

 

Try solving the following problems. Refer to math help for assistance, and if necessary, click on the "hint" buttons below for clues to working out the problem.

Example 2

A 5 kg box is sliding across the floor at 9 m/s. It slides over a rough section of the floor that exerts a frictional force of 15 Newtons for 3 meters. What is the velocity of the box after the 3 meters?

Hint 1 Identify all given information and verify that measurements are in the correct units.

mass = 5 kg     initial velocity = 9 m/s     final velocity = ?     Force = 15 Newtons     displacement = 3 meters     θ = 180° (friction is pointing directly opposite the displacement) Close

Hint 2 The correct forumula for this problem is W_net = ΔKE Close

Solution W_net = ΔKE
Fdcosθ = KE_f – KE_i
(15)(3)(cos180) = 1 2 (5)(v_f²) – 1 2 (5)(9²)
45(-1)= 2.5v_f² – (2.5)(81)
-45 = 2.5v_f² – 202.5
157.5 = 2.5v_f²
63 = v_f²
v_f = √63 = 7.94 m/s Close

 

Example 3

A 1250 kg boat is moving across a calm lake. The motor is pushing it forward with 5000 N of thrust, and the boat is feeling 3500 N of drag from the water. If it starts from rest, what is the final speed after it has moved 75 m?

Hint 1 This is a pretty complicated problem, to keep track of everything, it is best to draw a picture and identify all given information and verify that measurements are in the correct units.
There are two forces affecting the same displacement. Let's see how they line up.

Thrust:

Drag:


Mass = 1250 kg     initial velocity = 0m/s     final velocity = ?     displacement = 75m     Force₁(Thrust) = 5000 N     θ₁ = 0°     Force₂(Drag) = 3500 N     θ₂ = 180° Close

Hint 2 The correct formula for this problem is W_net = ΔKE Close

Solution Since we have more than one force, there are two different ways to look at the left side of this equation. We can figure out the work done by each force and add them together, or we can figure out the total force first, and get the work done by the total.

Wnet = ΣW
Wnet = F₁dcosθ₁ + F₂dcosθ₂
Wnet = (5000)(75)cosθ + (3500)(75)cos180
Wnet = 375000 – 262500
Wnet = 112500J

Wnet = F_net dcosθ
Wnet = (F₁ + F₂)dcosθ
Wnet = (5000 – 3500)(75)cosθ
Wnet = (1500)(75)
Wnet = 112500J

As you can see, either way we get the same result.

Now
    Wnet = ΔKE
112500 = ΔKE
112500 = KE_f – KE_i
112500 = 1 2 mv_f² – 1 2 mv _i²
112500 = 1 2 (1250)v_f² – 0
112500 = 625v_f²
      180 = v_f²
         v_f = 13.4 m/s Close