The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation t-x-3- where x is in meters and t is in seconds- The work done by the force in the first 6 seconds isA- 9 JB- 0 JC- 6 JD- 3 J

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Standard XII

Physics

Work Done When Force Is Varying

The displacem...

Question

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

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Solution

The correct option is C 0J
$x = (t - 3)^{2} \Rightarrow v = \frac{d x}{d t} = 2 (t - 3)$
at t = 0; $v_{1} = - 6 m / s$ and at t = 6 sec, $v_{2} = 6 m / s$
so, change in kinetic energy = $W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$

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The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation t=√x+3, where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

The correct option is C 0Jx=(t−3)2⇒v=dxdt=2(t−3) at t = 0; v1=−6 m/s and at t = 6 sec, v2=6 m/s so, change in kinetic energy = W=12mv22−12mv21=0

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

The correct option is C 0J
$x = (t - 3)^{2} \Rightarrow v = \frac{d x}{d t} = 2 (t - 3)$
at t = 0; $v_{1} = - 6 m / s$ and at t = 6 sec, $v_{2} = 6 m / s$
so, change in kinetic energy = $W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$