A truck weighing 1000 kg changes its speed from 36 km/h to 72 km/h in 2 minutes.

Calculate:

(i) the work done by the engine and

(ii) its power.

(g =10 m/ s^-2)

Given,

Mass = 1000 kg

$\text {Initial velocity (u) } = 36 km / h \\[0.5em] = \dfrac {36 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Initial velocity (u) } = 10ms^{-1} \\[0.5em]$

$\text {Final velocity (v) } = 72 km / h \\[0.5em] = \dfrac {72 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Final velocity (v) } = 20ms^{-1} \\[0.5em]$

From, equation of motion we get

$S = ( v^2 - u^2) / time \\[0.5em]$

(i) Work done = force x displacement

$W = \dfrac{1000 \times (20^2 -10^2)}{2} \\[0.5em] W = \dfrac{1000 \times (400 -100)}{2} \\[0.5em] W = 500 \times 300 \\[0.5em] \Rightarrow W = 150000J = 1.5 × 10^5J \\[0.5em]$

(ii) Power = work done / time taken

$Power = \dfrac{1.5 \times 10^5}{120} \\[0.5em] \Rightarrow Power = 1.25 × 10^3 W$