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 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 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 x of a particle of mass m kg moving in one dimension, under the action of a force, is related to the time t by the equation t=4x+3 where x is in meters and t is in seconds. The work done by the force in the first six seconds in joules is

The position (x) of a particle of mass 1 kg moving along X-axis at time t is given by (x=1/2 t^(2)) metre. Find the work done by force acting on it in time interval from t=0 to t=3 s .

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

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 ).

The displacement x of a particle at time t moving under a constant force is t=sqrtx+3 , x in metres, t in seconds. Find the work done by the force in the interval from t=0 to t=6 second.