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A point mass of 0.5 kg is moving along x-axis as x = t² + 2t, where, x is in meters and r is in seconds. Find the work done by all the forces acting on the body during the time interval [0, 2s]

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a

16J

b

4J

c

2J

d

8J

answer is D.

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Detailed Solution

Method : $W_{all} = ΔKE = \frac{1}{2} m (v_{2}^{2} - v_{0}^{2})$
$\begin{array}{l} v = \frac{dx}{dt} = 2 t + 2 = 2 (t + 1) \\ v_{2} = 6 m / s, v_{0} = 2 m / s \\ W_{all} = \frac{1}{2} (\frac{1}{2}) (6^{2} - 2^{2}) = 8 J \end{array}$
Method II : x = t² + 2t
$a = \frac{d^{2} x}{{dt}^{2}} = 2 m / s^{2} =$
So, F - m(l)= 1 N = constant
So, W = F.s = (1)(8-0) = 8 J

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