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Q.

The displacement x of a particle of mass 1kg moving in one dimension, under the action of a constant force is related to the time by the equation $t = \sqrt{x} + 3$ . Where x is in meters and t is in second. The workdone by the force in first 6 seconds is

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a

16 J

b

18 J

c

14 J

d

12 J

answer is B.

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Detailed Solution

$t = \sqrt{x} + 3$
$\begin{array}{l} \sqrt{x} = t - 3 \\ x = (t - 3)^{2} \\ x = t^{2} - 6 t + 9 \\ V = \frac{d x}{d t} = 2 t - 6 \end{array}$
$\begin{array}{l} a = \frac{d v}{d t} = 2 m s^{- 2} \\ A t t = 6 s, x = t^{2} - 6 t + q \\ = 6^{2} - 6 \times 6 + 9 \\ = 9 m \\ W = F S = M a x = 1 \times 2 \times 9 = 18 J \end{array}$

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