A dockworker applies a constant horizontal force of 89.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.5 m in a time of 5.50 s. A) What is the mass of the block of ice?B) If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20 s ?.

a) 90.9 kg

b) 22 m

Explanation:

(a)

    t = 5.50s

     t+1/2*a_x*t^2

PLAN:

find the acceleration, then m by ∑Fx = m*a_x

Let +x be the direction of the force

∑Fx =80 N

a_x=2()/t^2

      =0.880m/s^2

  m =∑Fx/a_x

      = 90.9 kg

For (b). first calculate the speed at the end of the period (4.5 s) of applied force (accelerated kinematics) Then use this ending velocity as the initial (and constant) in the second part of motion

(b) a_x = 0 and v_x is constant.

After the first 5.0 s

v_x=+a_x*t

     =4.40 m/s

This is now the initial speed for the motion of the next 4.20 s (a=0)

t+1/2*a_x*t^2

          = 22 m

note:

please verify the calculation

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.