A box of bananas weighing `40.0N `rests on a horizontal surface is `0.40 ` and the coefficient of kinetic friction is `0.20`
a. If no horizontal force is applied to the box and the box is at rest , how large is the friction force excered on the box ?
b. What is the magnitude of the friction forces if a monkey applies a horizontal force of `6.0 N ` to the box and the box is initial at rest?
c. What minimum horizontal force must the monkey apply to start the box in motion ?
d. What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been starts?
e. If the monkey applies a horizontal force of `18.0N ` What is the magnitude of the friction force and what is the box's acceleration ?

Explanation

a. If no horizontal force is applied , no friction is needed to keep the box in equlibrium
b. The maximum static friction force is
`mu_(s) N = mu_(s) W = (0.40)(40.0) = 16.0N`
so the box will not move and the friction force balances the applied force of `6.0 N` Hence friction` = 6 N`
c. The maximum friction force found the part (b) `16.0N`
d.From `f_(k) = mu_(k) N = (0.20)(40.0 N) = 8.0 N.`
e. The applied force is enough other to start the box moving or to keep it moving . The answer to part (d) is independent of speed (as long as the box is moving) so the friction force is `8.0 N` .The acceleration is `(F - f_(k)) //m = 2.5 m//s^(2)`