A box that weighs 100 lb is pushed along a surface by a variable force, F = 25 t^0.5 lb. The coefficient of friction between the box and the floor is 0.25 for both static and kinetic friction. What is the velocity of the box after 2 seconds? Assume the box starts from rest.

Impulse Action on Box

It is important first to note that t he block will not move until the force is large enough to overcome static friction. Thus, the exact time the box first moves needs to be determined.

The block will move when the force is equal to friction force to move. Summing the forces in the x-direction, gives,

     25 t^0.5 - f = 0
     μN = 25 t^0.5
      0.25 (100) = 25 t^0.5

     t = 1.0 s

The motion of the box will depend on the linear impulse between 1 and 2 seconds.

     
     (1.8856 - 2) - (0.6667 - 1) = 0.1242 v₂

v₂ = 1.763 ft/s