The mass of a body is 2.5 Kg. it is in motion and its velocity `upsilon`after time t is ` upsilon = (t^3)/(3) +(t^2)/(2) +1`
Calculate the force acting on the body at the time t =3 s.

The momentum of a body in two perpendicular direction at any time 't' are given by P_(x) =2t^(2) + 6 and P_(y) = (3t^(2))/2 + 3 . The force acting on the body at t= 2 sec is

The momentum of a body in two perpendicular direction at any time 't' are given by P_(x) =2t^(2) + 6 and P_(y) = (3t^(2))/2 + 3 . The force acting on the body at t= 2 sec is

The momenta of a body in two perpendicular directions at any time 't' are given by P_(x)=2t^(2) +6 and P_(y)=(3t^(2))/(2)+3 . The force acting on the body at t =2 sec is .

A particle of mass 3 kg is moving along x - axis and its position at time t is given by equation x=(2t^(2)+5)m . Work done by all the force acting on it in time interval t=0 to t=3s is

A particle of mass 3 kg is moving along x - axis and its position at time t is given by equation x=(2t^(2)+5)m . Work done by all the force acting on it in time interval t=0 to t=3s 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.