# The relation between the displacement x and the time t for a body of mass 2kg moving under the action of force is given by x=t3/3 , where x is in metre and t in second , calculate the work done by body in first 2 seconds.

Here,

Mass, m = 2 kg

Displacement is given as, x = t3/3

So, d(x)/dt = (3)(t2)/3 = t2

=> d2(x)/dt2 = 2t

=> a = 2t

This is the acceleration of the body.

Now,

For a small displacement ‘dx’ the work done is,

dW = F.dx

=> dW = (ma).d(t3/3)

=> dW = 2(2t)(3t2/3)

=> dW = 4t3 • 70

We obtain acceleration by double differentiating x wrt t

dx/dt = 3t2 /3 = t2 [this is velocity]

a = d2 x/ dt2 = 2t

F = ma = 2 kg x 2t = 4t newton

We see that force is variable. Let dW be small work done by variable force in displacing the body through dx

dx = t2 dt

dW = F. dx = 4t x t2 dt = 4t3 .dt

-

Net work done in the first 2 seconds = ∫4t3 .dt = t4

Now apply limits from 0 to 2

W = 24 - 04 = 16 J

• 46

We can also use Work Energy theorem.

Work Done = Change in Kinetic energy

For calculating KE we need velocity at time 0 and 2 s

v = dx/dt = t2

v at 0 sec = 0

v at 2 sec = 22 = 4 m/s

-

W = 1/2 m v2 = (1/2 ) x 2 x (4)2 = 16 J

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The answers obtained by both methods are same 16 J

• 133
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