An object having mass = 2kg  experience a forceF= 2t^{2}\widehat{i} in x - direction. Find work done by force in first 3 sec.

Answers (1)

Given,  m = 2 kg

F  = 2t^2 i

a = F/m =  2t^2/2 = t^2 i

a = \frac{\mathrm{d} v}{\mathrm{d} t} = t^{2}

\int_{u}^{v} \mathrm{d} v = \int_{0}^{t}\mathrm{d}t^{2}dt

v - u = \frac{t^{3}}{3}

Assuming initial velocity was zero, u = 0

v = \frac{t^{3}}{3}

v = \frac{\mathrm{d} x}{\mathrm{d} t} = \frac{t^{3}}{3}

\Rightarrow \mathrm{d}x = \frac{t^{3}}{3}.\mathrm{d}t

We know that, dw = F.dx

\mathrm{d}W = 2t^{2}\times \frac{t^{3}}{3}.\mathrm{d}t

\int_{0}^{W}\mathrm{d}W =\int_{0}^{t} 2t^{2}\times \frac{t^{3}}{3}.\mathrm{d}t

W = \frac{t^{6}}{9} \; J

Put t = 3 secs

W = \frac{3^{6}}{9} \; J

W = 81 J

 

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