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Physics

A man spends 6.4 kJ energy in displacing a body by 64 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.

Work, Energy & Power

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Answer

Given,

Work done = 6.4 kJ

S = 64 m

t = 2.5 s

(i) W = F x S

Force=work donedistanceF=6.4×10364F=100N\text{Force} = \dfrac{\text{work done}}{\text{distance}} \\[0.5em] F = \dfrac{6.4 \times 10^3}{64} \\[0.5em] \Rightarrow F = 100 N \\[0.5em]

(ii)

Power spent=work donetime takenP=6.4×1032.5P=2560W\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] P = \dfrac{6.4 \times 10^3}{2.5} \\[0.5em] \Rightarrow P = 2560 W \\[0.5em]

We know,

1 H.P=746 W1 W=1746 H.P2560 W=25607462560 W=3.43 H.P\text{1 H.P} = \text{746 W} \\[0.5em] \text{1 W}= \dfrac{1}{746} \text{ H.P} \\[0.5em] \therefore \text {2560 W} = \dfrac {2560}{746} \\[0.5em] \Rightarrow \text {2560 W} = \text{3.43 H.P}

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