Work-Energy Theorem Examples

A \(\text{1}\) \(\text{kg}\) brick is dropped from a height of \(\text{10}\) \(\text{m}\). 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.

Step 1: Determine what is given and what is required

• Mass of the brick: \(m=\text{1}\text{ kg}\).

• Initial height of the brick: \({h}_{i}=\text{10}\text{ m}\).

• Final height of the brick: \({h}_{f}=\text{0}\text{ m}\).

• 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 : Determine the brick's potential energy at \({h}_{i}\)

\begin{align*} E_p& = m·g·h\\ & = \left(1\right)\left(\text{9.8}\right)\left(10\right)\\ & = \text{98}\text{ J} \end{align*}

Step 3: Determine the work done on the brick

The brick had \(\text{98}\) \(\text{J}\) of potential energy when it was released and \(\text{0}\) \(\text{J}\) of kinetic energy. When the brick hit the ground, it had \(\text{0}\) \(\text{J}\) of potential energy and \(\text{98}\) \(\text{J}\) of kinetic energy. Therefore \({E}_{k,i}=\text{0}\text{ J}\) and \({E}_{k,f}=\text{98}\text{ J}\).

From the work-energy theorem:

\begin{align*} W_{\text{net}}& = \Delta E_k\\ & = {E}_{k,f}-{E}_{k,i}\\ & = 98-0\\ & = \text{98}\text{ J} \end{align*}

Hence, \(\text{98}\) \(\text{J}\) of work was done on the brick.

The driver of a \(\text{1 000}\) \(\text{kg}\) car travelling at a speed of \(\text{16.7}\) \(\text{m·s$^{-1}$}\) applies the car's brakes when he sees a red light. The car's brakes provide a frictional force of \(\text{8 000}\) \(\text{N}\). Determine the stopping distance of the car.

Step 1: Determine what is given and what is required

We are given:

• mass of the car: \(m = \text{1 000}\text{ kg}\)

• speed of the car: \(v = \text{16.7}\text{ m·s$^{-1}$}\)

• frictional force of brakes: \(\vec{F} = \text{8 000}\text{ N}\)

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:

\(E_k=\frac{1}{2}m{v}^{2}\)

This energy is equal to the work done by the brakes. We have the force applied by the brakes, and we can use:

\(W=F \Delta x \cos \theta\)

to determine the stopping distance.

Step 3: Determine the kinetic energy of the car

\begin{align*} E_k& = \frac{1}{2}m{v}^{2}\\ & = \frac{1}{2}\left(1000\right){\left(\text{16.7}\right)}^{2}\\ & = \text{139 445}\text{ J} \end{align*}

Step 4: Determine the work done

Assume the stopping distance is \({\Delta x}_{0}\). Since the direction of the applied force and the displacement are in opposite directions, \(\theta = 180°\). Then the work done is:

\begin{align*} W& = F \Delta x \cos \theta\\ & = \left(8000\right)\left( \Delta x_0 \right) \cos (180) \\ & = \left(8000\right)\left( \Delta x_0 \right)(-\text{1}) \\ & = \left(-\text{8 000}\right)\left( \Delta x_0 \right) \end{align*}

Step 5: Apply the work-enemy theorem

The change in kinetic energy is equal to the work done.

\begin{align*} \Delta E_k& = W_{\text{net}}\\ {E}_{k,f}-{E}_{k,i}& = \left(-8000\right)\left({\Delta x_0}\right)\\ \text{0}-\text{139 445} & = \left(-8000\right)\left({\Delta x_0}\right)\\ \therefore \Delta x_0 & = \frac{139445}{8000}\\ & = \text{17.4}\text{ m} \end{align*}

Step 6: Write the final answer

The car stops in \(\text{17.4}\) \(\text{m}\).