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 .

The displacement of a body of mass 2 kg veries 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 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

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

The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as x(m)=3t^(2)+4t+5 Here, t is expressed in second. What is the work done by the force in first 2 seconds?

A force F=a+bx acts on a particle in x-direction, where a and b are constants. Find the work done by this force during the displacement from x_1 to x_2 .