The distance x covered by a body of 2 kg under the action of a force is related to time t as `x=t^(2)//4`. What is the work done by this force in 2 seconds ?

The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as x(m)=3t^(2)+4t+5 Here, t is expressed in second. What is the work done by the force in first 2 seconds?

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

The displacement x in meters of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in seconds by the equation t=sqrtx+3 , , the work done by the force (in J) in first six seconds is :-

A body covers a distance of 4m under the action of force F=(17-4x)N where x is the distance in metres. The work done by the force is

The displacement x in meter of a particle of mass mkg moving in one dimenstion under the action of a force is related to the time t in second by the equation x=(t-3)^2 . The work done by the force (in joules) in first six seconds is

A body covers a distance of 4m under the action of force F=(17-4x)N where x is the distance in metres.The work done by the force is

A body covers a distance of 4m under the action of force F=(17-4x)N where x is the distance in metres.The work done by the force is

A body covers a distance of 4m under the action of force F = (17 – 4x)N where x is the distance in metres. The work done by the force is

The relation between the displacement x and the time t for a body of mass 2 kg moving under the action of a force is geven by x=t^(3)//3 , where x is in metre and t in second ,Calculate the ork done by the body in first 2 seconds.