A block of mass `2kg` at rest at origin is acted upon by a force `F=(x^(3)-3x+1)N` , where `x` is in metre. The speed of the block at `x=2m` is

A block of mass m at rest is acted upon by a force F for a time t. The kinetic energy of block after time t is

A particle of mass m is acted upon by a force F= t^2-kx . Initially , the particle is at rest at the origin. Then

A box of mass m=2 kg resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle when force ceases to act.

A block of the mass of 1 kg is moving on the x-axis . A force F acting on the block is shown . The velocity of the block at time t = 2 s is - 3 m s^(-1) what is the speed of the block at time t = 4 s?

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an axis is applied to the block. The force is given by vecF=(1-x^(2))hatiN , where x is in meters and the initial position of the block is x=0 the maximum kinetic energy of the block is (2)/(n)J in between x=0 ad x=2 m find the value of n .

A small block of mass 1kg is at rest on a parabolic surface given by y = x^(2) . The coefficient of friction between the block and the surface is enough to prevent slipping.If position of block on surface is( 2m, 4m) then the normal force then the surface exerts at this position is

A 1.5-kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by vecF=(4-x^2)veciN , where x is in meter and the initial position of the block is x=0 . The maximum kinetic energy of the block between x=0 and x=2.0m is A.2.33 J B.8.67 J C.5.33 J D.6.67 J

A block of mass 3kg is at rest on a rough inclined plan as shown in the Figure. The magnitude of net force exerted by the surface on the block will be

A small block of mass 1 kg is at rest on a parabolic surface given by y = x^2 . The coefficient of friction between the block and the surface is enough to prevent slipping. If position of block on surface is (2 m, 4 m), then the normal force that the surface exerts at this position is

A block of mass 20 kg is acted upon by a force F=30N at an angle 53^@ with horizontal in downward direction as shown. The coefficient of friction between the block and the forizontal surface is 0.2 . The friction force acting on the block by the ground is (g=10(m)/(s^2) )