Students will learn that work is simply the transfer of energy into or out of a system. Students will learn how to calculate the work done and how to incorporate it into energy conservation.

### Key Equations

\begin{align*}W = F_{x} \Delta x = Fd \ \cos \circleddash\end{align*}

\begin{align*}W =\end{align*}

### Guidance

When an object moves in the direction of an applied force, we say that the force does **work** on the object. Note that the force may be slowing the object down, speeding it up, maintaining its velocity --- any number of things. In all cases, the net work done is given by this formula:

\begin{align*} W = \vec{F}\cdot \vec{d} = \vec{F}\cdot \Delta \vec{x} && \text {Work is the dot product of force and displacement.} \end{align*}

In other words, if an object has traveled a distance \begin{align*} d \end{align*}

Here the net work done on the object by the force will be \begin{align*}F d \cos \theta\end{align*}

#### Example 1

A 1kg ball has been attached to a 2m string and is at rest on a frictionless surface. If you exert a constant force of 10N on the string and pull the ball over the course of 5m and then begin spinning the ball in a circle, after 3 revolutions, what is the total amount of work you have done on the ball?

##### Solution

Since the centripetal force you exert on the ball in order to make it spin is perpendicular to the ball's path, you do not work on the ball while spinning it in a circle. Therefore, the only work you do on the ball is when you are pulling it in a straight line.

\begin{align*} W&=Fd\\ W&=10\text{N} * 5\text{m}\\ W&=50\text{J}\\ \end{align*}

#### Example 2

A block of mass 5kg is sliding down a ramp inclined at 45 degrees. If the coefficient of kinetic friction between the ramp and the block is 0.3, how much work does the force of friction do as the block slides 3m down to the bottom of the ramp.

##### Solution

In order to find the work done by friction, we first want to find out the magnitude of the force of friction.

\begin{align*} f&=\mu_kN && \text{start with the equation for the force of friction}\\ f&=\mu_kmg\cos(30) && \text{substitute the y-component of the weight of the block for the normal force}\\ \end{align*}

Now that we have the magnitude of the force of friction, we can plug that into the equation for work.

\begin{align*} W&=Fd\\ W&=\mu_kmgcos(45)\\ W&=0.3*5\text{kg}*9.8\text{m/s}^2*\cos(45)\\ W&=10.4J \end{align*}

### Watch this Explanation

### Explore More

- You slide down a hill on top of a big ice block as shown in the diagram. Your speed at the top of the hill is zero. The coefficient of kinetic friction on the slide down the hill is zero \begin{align*}(\mu_k = 0)\end{align*}
(μk=0) . The coefficient of kinetic friction on the level part just beneath the hill is \begin{align*}0.1 (\mu_k = 0.1)\end{align*}0.1(μk=0.1) .- What is your speed just as you reach the bottom of the hill?
- How far will you slide before you come to a stop?

- Marciel is at rest on his skateboard (total mass \begin{align*}50\;\mathrm{kg}\end{align*}
50kg ) until he catches a ball traveling with a speed of \begin{align*}50\;\mathrm{m/s}\end{align*}50m/s . The baseball has a mass of \begin{align*}2\;\mathrm{kg}\end{align*}2kg . What percent of the original kinetic energy is transferred into heat, sound, deformation of the baseball, and other non-mechanical forms when the collision occurs? - Investigating a traffic collision, you determine that a fast-moving car (mass \begin{align*}600\;\mathrm{kg}\end{align*}
600kg ) hit and stuck to a second car (mass \begin{align*}800\;\mathrm{kg}\end{align*}800kg ), which was initially at rest. The two cars slid a distance of \begin{align*}30.0\;\mathrm{m}\end{align*}30.0m over rough pavement with a coefficient of friction of \begin{align*}0.6\end{align*}0.6 before coming to a halt. What was the speed of the first car? Was the driver above the posted \begin{align*}60\;\mathrm{MPH}\end{align*}60MPH speed limit? - Force is applied in the direction of motion to a \begin{align*}15.0\;\mathrm{kg}\end{align*}
15.0kg cart on a frictionless surface. The motion is along a straight line and when \begin{align*}t = 0\end{align*}t=0 , then \begin{align*}x = 0\end{align*}x=0 and \begin{align*}v = 0\end{align*}v=0 . (The displacement and velocity of the cart are initially zero.) Look at the following graph:- What is the change in momentum during the first \begin{align*}5\end{align*}
5 sec? - What is the change in velocity during the first \begin{align*}10\end{align*}
10 sec? - What is the acceleration at \begin{align*}4\end{align*}
4 sec? - What is the total work done on the cart by the force from \begin{align*}0 - 10\end{align*}
0−10 sec? - What is the displacement after \begin{align*}5\end{align*}
5 sec?

- What is the change in momentum during the first \begin{align*}5\end{align*}
- Force is applied in the direction of motion to a \begin{align*}4.00\;\mathrm{kg}\end{align*}
4.00kg cart on a frictionless surface. The motion is along a straight line and when \begin{align*}t = 0, v = 0\end{align*}t=0,v=0 and \begin{align*}x = 0\end{align*}x=0 . look at the following graph:- What is the acceleration of the cart when the displacement is \begin{align*}4\;\mathrm{m}\end{align*}
4m ? - What work was done on the cart between \begin{align*}x = 3\;\mathrm{m}\end{align*}
x=3m and \begin{align*}x = 8\;\mathrm{m}\end{align*}x=8m ? - What is the total work done on the cart between \begin{align*}0 -10\;\mathrm{m}\end{align*}
0−10m ? - What is the speed of the cart at \begin{align*}x = 10\;\mathrm{m}\end{align*}
x=10m ? - What is the impulse given the cart by the force from \begin{align*}1 - 10\;\mathrm{m}\end{align*}
1−10m ? - What is the speed at \begin{align*}x = 8\;\mathrm{m}\end{align*}
x=8m ? - How much time elapsed from when the cart was at \begin{align*}x = 8\end{align*}
x=8 to when it got to \begin{align*}x = 10\;\mathrm{m}\end{align*}x=10m ?

- What is the acceleration of the cart when the displacement is \begin{align*}4\;\mathrm{m}\end{align*}

#### Answers to Selected Problems

- a. \begin{align*}10 \;\mathrm{m/s}\end{align*}
10m/s b. \begin{align*}52\;\mathrm{m}\end{align*} - \begin{align*}96\end{align*}%
- \begin{align*}43.8 \;\mathrm{m/s}\end{align*}
- .
- .