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PhysicsW.E.P.Work kinetic energy theoremEasy2 minQB
PhysicsEasysingle choice

A force acts on a2 kgobject so that its position is given as a function of time asx=3t2+5.What is the work done by this force in first5seconds?

Options:

Answer:
D
Solution:

Here, the position of the object is x=3t2+5, therefore the velocity of the object will be,

v=dxdt=d3t2+5dt

v=6t+0

From 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=KEt=5 s-KEt=0 s
W=12×2×6×52-12×2×6×02
W=900 J.

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Work kinetic energy theorem
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