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A force F=(3ti^+5j^)N acts on a body due to which its position varies as S=(2tsqi^-5j^).Work done by this force in first two seconds is

A force F=(3ti^+5j^)N acts on a body due to which its position varies as S=(2tsqi^-5j^).Work done by this force in first two seconds is

Grade:11

2 Answers

Rohana
38 Points
6 years ago
P
Jai krishan
15 Points
5 years ago
we know that
∫dW = ∫f.ds
f = 3Ti+5j
ds = 2t2-5j
f.ds = f.ds/dt.dt
ds/dt = 4Ti
∫dW = ∫ (3Ti+5j).(4Ti).dt
W = [4T3]
for t=0 to t=2
W = 32 J

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