First slide

Work

Question

under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x= $\frac{t^{3}}{3}$ , where x is in metre and t is in second. The work done by the force in the first two seconds is

Moderate

A

1600 Joule
B

160 Joule
C

16 Joule
D

1.6 Joule

Solution

$\begin{matrix} g i v e n p o s i t i i o n x = \frac{t^{3}}{3} \Rightarrow V e l o c i t y = \frac{d x}{d t} = t^{2} \\ v e l o c i t y o f t h e b o d y v = t^{2} \\ W o r k d o n e = c h a n g e i n K E \\ W o r k = \frac{1}{2} m v^{2}; s u b s t i t i t e f o r v e l o c i t y \\ W = \frac{1}{2} m {(t^{2})}^{2}; s u b s t i t u t e t = 2; m = 2 (g i v e n) \\ W = \frac{1}{2} \times 2 \times 2^{4} \\ W = 16 J = w o r k d o n e i n f i r s t 2 s e c o n d \end{matrix}$

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