A man spends 7.4 kJ energy in displacing a body by 74 m in the direction in which he applies force, in 2.5 s.

Calculate:

(i) the force applied and

(ii) the power spent (in H.P) by the man.

Given,

Work done = 7.4 kJ

S = 74 m

t = 2.5 s

(i) W = F x S

$\text{Force} = \dfrac{\text{work done}}{\text{distance}} \\[0.5em] F = \dfrac{7.4 \times 10^3}{74} \\[0.5em] \Rightarrow F = 100 N \\[0.5em]$

(ii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] P = \dfrac{7.4 \times 10^3}{2.5} \\[0.5em] \Rightarrow P = 2960 W \\[0.5em]$

We know,

$\text{1 H.P} = \text{746 W} \\[0.5em] \text{1 W}= \dfrac{1}{746} \text{ H.P} \\[0.5em] \therefore \text {2960 W} = \dfrac {2960}{746} \text{ H.P} \\[0.5em] \Rightarrow \text {2960 W} = \text{3.97 H.P}$