Worked Examples for Work

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

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:

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

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

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:

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

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:

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

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:

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

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)