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

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 .

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 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