First slide

Work done by Diff type of forces

Question

The displacement x in metres of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in seconds by the equation t = $\sqrt{x} + 3$ , the work done by the force (in J) in first six seconds is:

Moderate

A

18 J
B

zero
C

9 J
D

36 J

Solution

Here t = $\sqrt{x} + 3 or x = {(t - 3)}^{2} = t^{2} - 6 t + 9$
$v = \frac{dx}{dt} = 2 t - 6$
at t = 0, v = 2 $\times$ 0 -6 = -6
at t = 6 s, v = 2 $\times$ 6-6 =+6
Initial and final KE are same hence no work is done.

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