A block of mass 2 kg is kept at origin at t = 0 and is having velocity `4sqrt5m//s` in positive x - direction. The only force on it is a conservative and its potential energy is defined as `U=-x^(3)+6x^(2)+15` (SI units). Its velocity when the force acting on it is minimum (after the time t = 0 ) is

A block of mass 2 kg is having velocity 4sqrt(5) ms^(-1) in the positive x - direction at the origin. The only force acting on it is F = (3x^2 - 12 x)N . Its velocity when it is at x = 2 m is

A block of mass 2 kg is having velocity 4sqrt(5) ms^(-1) in the positive x - direction at the origin. The only force acting on it is F = (3x^2 - 12 x)N . Its velocity when it is at x = 2 m is

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.

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

A particle moves along x-axis and its displacement at any time is given by x(t) = 2t^(3) -3t^(2) + 4t in SI units. The velocity of the particle when its acceleration is zero is

A particle moves along X-axis and its displacement at any time is given by x(t) = 2t^(3) -3t^(2) + 4t in SI units. The velocity of the particle when its acceleration is zero, is

A force F of constant magnitude acts on a block of mass 'm' and t=0,initially along positive x-direction. The direction of force F starts rotating with a constant angular velocity omega in anticlockwise sense. Then the magnitude of velocity of the block as a function of time 't' is ( initial block was at rest )