When a force acts on a body of mass m," its position "x" varies with time "t" as `x=at^(4)+bt+c`" where "a,b" and "c are constants.The work done by the force during the first second of the motion is

A force of 5 N acts on a 15 kg body initially at rest. The work done by the force during the first second of motion of the body 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 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 .

The displacement of a body of mass 2 kg varies with time t as S = t^(2) + 2t , where S is in seconds. The work done by all the forces acting on the body during the time interval t = 2s to t = 4s 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