Force F is applied on upper pulley. If F = 30 t where t is time in second. Find the time when `m_(1)` loses contact with floor : (Take `g = 10 m//s^(2)`)

In the figure shown m_(1) = 1 kg , m_(2) = 2 kg , pulley is ideal . At t = 0 , both masses touches the ground and string is tout . A force F = 2t is applied to pulley ( t is in second ) then m_(1) is lifted off the ground at time ( g = 10 m//s^(2))

If v=(t^(2)-4t+10^(5)) m/s where t is in second. Find acceleration at t=1 sec.

Force applied on a body is given by ltbygt F=(3t^(2)-2t+10) N where t is in seconds. Find impulse imparted in t=0 to t=2 sec.

A block of mass 5 kg is kept on a rough horizontal surface. The coefficient of friction for the contact force block and surface are mu_(s) = 0*1 and mu_(k) = 0*9 respectively . A horizontal force F given by F 0*5t is applied at t = 0 . Here t represent time in seconds . Determine the force of friction and acceleration of the block at (i) t = 1s (ii) t= 10 s (iii) t = 12 s

A force is given by F= at + bt^(2) where , t is the time .The dimensione of a and b are

A force F is applied horizontally on mass m_(1) as shown in figure. Find the contact force between m_(1) and m_(2)

If force (F) is given by F = Pt^(-1) + alpha t, where t is time. The unit of P is same as that of

A block of 7 kg is placed on a rough horizontal surface and is pulled through a variable force F(in N) =5t, where 't' is time in second, at an angle of 37^(@) with the horizontal as shown in figure. The coefficient of static friction of the block with the surface is one. if the force starts acting at t=0 s, find the time at which the block starts to slide. (Take g=10 m//s^(2) )

A block of mass 10 kg is kept on ground. A vertically upward force F=(20 t)N , where t is the time in seconds starts on it at t=0 . (a) Find the time at which the normal reaction acting on the block is zero. (b) The height of the block from ground at t=10 sec.