Question

A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of work done by the car? The total mass of the car, including its passengers, is 2.5 × 103 kilograms.

Answer 1

The work done is $2.8125 \times 10^{5} \mathrm{J}$

Work Done = Change in Kinetic Energy (ΔKE)

Explanation

In first 1 hour it travels 72 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}$

or, Initial Velocity (u) = 20 m/s

Similarly for the next hour it covers 90 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}$

or, Final Velocity (v) = 20 m/s

Work done = Change in Kinetic Energy (ΔKE)

Work done = ΔKE = $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$

ΔKE = $\frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)$

= $\frac{2500 \times(625-400)}{2}=\frac{2500 \times 225}{2}=\frac{562500}{2}$= 281250 joule  

= $2.8125 \times 10^{5} \mathrm{J}$