The displacement x of a body of mass 1 kg on a horizontal smooth surface as a function of time t is given by `x = (t^4)/4`. The work done in the first second is

The displacement of a particle of mass 1kg on a horizontal smooth surface is a function of time given by x=1/3t^3 . Find out the work done by the external agent for the first one second.

The displacement of a particle of mass 1kg on a horizontal smooth surface is a function of time given by x=1/3t^3 . Find out the work done by the external agent for the first one second.

Displacement x ( in meters ) , of a body of mass 1 kg as a function of time t, on a horizontal smooth surface , is given as x = 2 t^2 Then work done in the first one second by the external force is

Displacement x ( in meters ) , of a body of mass 1 kg as a function of time t, on a horizontal smooth surface , is given as x = 2 t^2 Then work done in the first one second by the external force is

Under the action of a force a 2 kg body moves such that its position x in meters as a function of time t is given by x=(t^(4))/(4)+3. Then work done by the force in first two seconds is

Under the action of a force a 2 kg body moves such that its position x in meters as a function of time t is given by x=(t^(4))/(4)+3. Then work done by the force in first two seconds is

Under the action of force 2 kg body moves such that its position 'x' varies as a function of time t given by : x = t^(2)//2 . The work done by the force in the first 5 seconds is

The displacement x (in m ), of a patticle of mass m (in kg ) is related to the time t (in second) by t = sqrt(x) +3 . Find the work done in first six second. (in mJ ).