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 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 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 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 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 ?

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 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 .