The displacement x of a particle at time t moving under a constant force is t- x -3- x in metre- t in second- Find the work done by the interval from t-0 to t-6 s-

Given, t=√( x ) +3 ⇒ x=( t 3 ) nbsp;^ 2 ⇒ v=2( t 3 ) nbsp; nbsp; nbsp; nbsp; ( ∵ v= dx dt nbsp;) At t=0, v= 6 ms ^ ...

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Standard XII

Physics

Gravtiational PE

The displacem...

Question

The displacement $x$ of a particle at time $t$ moving under a constant force is $t = \sqrt{x} + 3$ , $x$ in metre, $t$ in second. Find the work done by the interval from $t = 0$ to $t = 6 s$ .

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Solution

Given, $t = \sqrt{x} + 3$
$\Rightarrow x = {(t - 3)}^{2}$
$\Rightarrow v = 2 (t - 3)$ $(∵ v = \frac{d x}{d t})$
At $t = 0, v = - 6 {m s}^{- 1}$
At $t = 6 s, v = 6 {m s}^{- 1}$
Work done $=$ Change in kinetic energy $= 0$ .

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The displacement x of a particle at time t moving under a constant force is t=√x+3, x in metre, t in second. Find the work done by the interval from t=0 to t=6s.

Given, t=√x+3⇒x=(t−3)2⇒v=2(t−3) (∵v=dxdt)At t=0,v=−6ms−1At t=6s,v=6ms−1Work done = Change in kinetic energy =0.

The displacement $x$ of a particle at time $t$ moving under a constant force is $t = \sqrt{x} + 3$ , $x$ in metre, $t$ in second. Find the work done by the interval from $t = 0$ to $t = 6 s$ .

Given, $t = \sqrt{x} + 3$
$\Rightarrow x = {(t - 3)}^{2}$
$\Rightarrow v = 2 (t - 3)$ $(∵ v = \frac{d x}{d t})$
At $t = 0, v = - 6 {m s}^{- 1}$
At $t = 6 s, v = 6 {m s}^{- 1}$
Work done $=$ Change in kinetic energy $= 0$ .