Work Energy Power

JEE Main level, single correct, medium

116 of 183

Question 116 of 183

A force acts on a 2kg2kg object so that its position is given as a function of time as x=3t2+5.x=3{t}^{2}+5. What is the work done by this force in first 55 seconds?

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Answer: (D) 900J900J

Here, the position of the object is x=3t2+5x=3{t}^{2}+5, therefore the velocity of the object will be,
v=dxdt=d(3t2+5)dtv=\dfrac{dx}{dt}=\dfrac{d\left(3{t}^{2}+5\right)}{dt}
v=6t+0v=6t+0From the work-energy theorem, all the work done by the force acting on it will be equal to the change in its kinetic energy, i.e.,
W=KE(t=5s)−KE(t=0s)W=KE\left(t=5s\right)-KE\left(t=0s\right)
W=12×2×(6×5)2−12×2×(6×0)2W=\dfrac{1}{2}\times 2\times {\left(6\times 5\right)}^{2}-\dfrac{1}{2}\times 2\times {\left(6\times 0\right)}^{2}
W=900JW=900J.

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