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)`)

Force applied on a body is given by F=(3t^(2)-2t+30) N where t is in seconds. Find impulse imparted in t=0 to t=2 sec.

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

In the figure shown m_(1) = 1 kg , m_(2) = 2 kg , pulley is ideal . At t = 0 , both msses 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))

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.

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

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.

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