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
Force=work donedistanceF=7.4×10374⇒F=100N\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]Force=distancework doneF=747.4×103⇒F=100N
(ii)
Power spent=work donetime takenP=7.4×1032.5⇒P=2960W\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]Power spent=time takenwork doneP=2.57.4×103⇒P=2960W
We know,
1 H.P=746 W1 W=1746 H.P∴2960 W=2960746 H.P⇒2960 W=3.97 H.P\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}1 H.P=746 W1 W=7461 H.P∴2960 W=7462960 H.P⇒2960 W=3.97 H.P
