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

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 foece, 1 kg body moves such that its position x as a function of time t is given by x=(t^(3))/(3), x is meter. Calculate the work done (in joules) by the force in first 2 seconds.

Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x =(t^(3))/(3) where x is in meter and t in second. The work done by the force in the first two seconds is .

Under the action of a force, a 1 kg body moves, such that its position x as function of time t is given by x=(t^(3))/(2). where x is in meter and t is in second. The work done by the force in fiest 3 second is

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 ?

Under the action of a force a block of mass 2 kg is moved in such a way that its position is given by x=(t^(3))/3 metre. What is the work done by the force in the first two 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

A point moves such that its displacement as a function of times is given by x^(2)=t^(2)+1 . Its acceleration at time t 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