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Standard XII

Physics

Work Energy Theorem Application

A force acts ...

Question

A force acts on a $30$ g particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metre and t is in second. The work done during the first $4$ second is?

5.28

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Solution

The correct option is A $5.28$ J
$x = 3 t - 4 t^{2} + t^{3}$
$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$
According to work energy theorem
$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$
$= \frac{1}{2} 0.03 (19^{2} - 3^{2}) = 5.28$

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A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

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A force acts on a 30g particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3, where x is in metre and t is in second. The work done during the first 4 second is?

The correct option is A 5.28Jx=3t−4t2+t3 v=dxdt=3−8t+3t2According to work energy theoremW=12m(v24−v20) =120.03(192−32)=5.28

A force acts on a $30$ g particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metre and t is in second. The work done during the first $4$ second is?

The correct option is A $5.28$ J
$x = 3 t - 4 t^{2} + t^{3}$
$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$
According to work energy theorem
$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$
$= \frac{1}{2} 0.03 (19^{2} - 3^{2}) = 5.28$