# Worked Examples for Work

Show/Hide Sub-topics (Work, Energy & Power | O Level)

These worked examples will help to solidify your understanding towards the concept of work and work done.

## Worked Example 1: Pushing A Box

### A boy pushes a box across a rough horizontal floor. He exerts 5 N to move it by 2 m. What is the work done by the boy?

\begin{aligned} \text{Work done by boy} &= f \times d \\ &= 5 \, \text{N} \times 2 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

## Worked Example 2: Pulling A Wagon

### A boy pulls a 1 kg toy wagon along a smooth (i.e. no friction) horizontal floor over a distance of 5 m. If the speed of the wagon increases at a constant rate of $2 \, \text{m s}^{-2}$, what is the work done by the boy?

The net resultant force on the toy wagon can be given by Newton’s Second Law: $f_{\text{resultant}} = ma$. Since the floor is smooth (no friction), the net resultant force on the toy wagon is just the force exerted by the boy.

Hence,

\begin{aligned} \text{Net Resultant Force} &= \text{Force exerted by boy} \\ \text{Force exerted by boy} &= m \times a \\ &= 1 \, \text{kg} \times 2 \, \text{m s}^{-2} \\ &= 2 \, \text{N} \end{aligned}

The work done by the boy will just be:

\begin{aligned} \text{Work done by boy on wagon} &= \text{force} \times \text{dist. moved in direction of force} \\ &= 2 \, \text{N} \times 5 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

## Worked Example 3: Raising a Box

### A 5 kg box is raised 50 m from its original position. What is the gain in gravitational potential energy? Assume gravitational field strength = $10 \, \text{N kg}^{-1}$. How much work is needed to raise the box to its new position?

Note: Assume that no energy is lost to air resistance.

\begin{aligned} E_{p} &= mgh \\ &= 5 \, \text{kg} \times 10 \, \text{N kg}^{-1} \times 50 \, \text{m} \\ &= 2500 \, \text{J} \end{aligned}

## Worked Example 4: Dropping A Ball

### A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level).

1. What is it’s initial gravitational potential energy?
2. Determine its velocity just before it hits the ground.

Note: Assume that no energy is lost to air resistance.

Part a:

Taking the ground level to be the reference level,

(For gravitational potential energy, the relevant formula is:)

\begin{aligned} E_{p} &= mgh \\ &= \left( 0.5 \, \text{kg} \right) \left( 10 \, \text{N kg}^{-1} \right) \left( 10 \, \text{m} \right) \\ &= 50 \, \text{J} \end{aligned}

Part b:

From the Principle of Conservation of Energy, the initial gravitational potential energy will be converted to its final kinetic energy. (I.e. Gain in kinetic energy = loss in gravitational potential energy)

For kinetic energy, the relevant formula is:

\begin{aligned} E_{k} &= \frac{1}{2} m v^{2} \\ 50 \, \text{J} &= \frac{1}{2} \left( 0.5 \, \text{kg} \right) v^{2} \\ v &= 14.1 \, \text{m s}^{-1} \end{aligned}

## Worked Example 5: Dropping A Ball With The Effects of Air Resistance

### A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level). There is an energy loss of $10 \, \text{J}$ due to air resistance. Determine its velocity just before it hits the ground.

You can obtain the gravitational potential energy of the ball from Worked Example 4, which is $50 \, \text{J}$.

By the Principle of Conservation of Energy, the initial gravitational potential energy will be converted to final kinetic energy and energy lost due to air resistance. Hence, we have:

\begin{aligned} E_{\text{initial}} &= E_{\text{final kinetic energy}} +\text{Energy Lost} \\ 50 \, \text{J} &= E_{\text{final kinetic energy}} + 10 \, \text{J} \\ E_{\text{final kinetic energy}} &= 50 \, \text{J}-10 \, \text{J} \\ &= 40 \, \text{J} \end{aligned}

Using the formula for kinetic energy, the velocity is:

\begin{aligned} E_{\text{final kinetic energy}} &= \frac{1}{2}mv^{2} \\ 40 \, \text{J} &= \frac{1}{2} \left( 0.5 \text{kg} \right) v^{2} \\ v &= 12.6 \, \text{m s}^{-1} \end{aligned}

## Worked Example 6: The Ball Rebounds

### A ball with a mass of 500 g is thrown vertically downwards from a height of 10 m with a velocity of $5.0 \, \text{m s}^{-1}$. It hits the ground and bounces. It then rises to a maximum height of 8.0 m. Determine the energy loss due to the bounce.

Assume that no energy is lost due to air resistance.

From the Principle of Conservation of Energy, the initial kinetic energy and initial gravitational potential energy is equals to the final kinetic energy + final gravitational potential energy + energy lost due to bounce. Let’s put this into an equation form:

\begin{aligned} E_{\text{k, initial}} + E_{\text{p, initial}} &= E_{\text{k, final}} + E_{\text{p, final}} + \text{Energy lost} \\ \frac{1}{2}mv^{2} + mgh_{\text{initial}} &= 0 + mgh_{\text{final}} + \text{Energy lost} \\ \frac{1}{2}\left( 0.5 \right) \left( 5^{2} \right) + \left( 0.5 \right) \left( 10 \right) \left( 10 \right) &= \left( 0.5 \right) \left( 10 \right) \left( 8 \right) + \text{Energy lost} \\ 56.25 \, \text{J} &= 40 \, \text{J} + \text{Energy lost} \\ \text{Energy lost} &= 16.25 \, \text{J} \\ \text{Energy lost} &= 16.3 \, \text{J} \, \left(3 \, \text{s.f.} \right) \end{aligned}

## Worked Example 7: Work Done By Car Engine

### The engine of a car exerts a constant force of 10 kN. The car is able to accelerate constantly from rest to a speed of $30 \, \text{m s}^{-1}$ in 10 seconds. Determine the work done by the engine of the car and the kinetic energy of the car at the end of the 10 seconds.

From what you have learnt in Kinematics,

\begin{aligned} v &= u + at \\ a &= \frac{\left( 30-0 \right)}{10} \\ &= 3 \, \text{m s}^{-1} \end{aligned}

With the given engine force and calculated acceleration, we can find the mass of car by employing the use of Newton’s Second Law. (Note: The engine force is the net resultant force acting on the car. Can you see why?)

\begin{aligned} F &= ma \\ m &= \frac{F}{a} \\ &= \frac{10000}{3} \\ &= 3330 \, \text{kg} \end{aligned}

The distance travelled by the car in 10 seconds is given by:

\begin{aligned} d &= \frac{1}{2} \left( v + u \right) \times t \\ &= 150 \, \text{m} \end{aligned}

The work done by the engine of the car will be:

\begin{aligned} W &= F \times d \\ &= 10000 \, \text{N} \times 150 \, \text{m} \\ &= 1 500 000 \, \text{J} \\ &= 1.5 \times 10^{6} \, \text{J} \end{aligned}

The kinetic energy of the car at the end of 10 seconds will be given by:

\begin{aligned} E_{k} &= \frac{1}{2} mv^{2} \\ &= \frac{1}{2} \left( 3330 \, \text{kg} \right) \left( 30 \, \text{m s}^{-1} \right)^{2} \\ &= 1.5 \times 10^{6} \end{aligned}

We notice that the work done by the engine of the car is the same as the kinetic energy of the car. This shows that all the work done by the engine of the car is converted into the kinetic energy of the car. (which is true if there is no dissipative forces acting on the car)