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 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 .

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 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 metre and t in second. The work done by the force in the first two seconds is .