Question

Question 6 1 pts A box of bananas weighing 40.ON rests on a horizontal surface. The coefficient of static friction (us) between the box and the surface is 0.40 while the coefficient of kinetic friction (uk) is 0.20. If a monkey applies a constant horizontal force of 6.0 N, and the box is initially at rest, what is the magnitude of the friction force 1 minutes after the monkey starts pushing? 16.0N 8.0 N Unable to determine O 0.ON 18.0N 6.0 N 40.0N Another value not shown here o e

Answer 1

friction of ng

let

$F_{app}=$force- applied by the monkey in right directon

$\mu_{s}=0.4$= coefficient of static friction

$\mu_{k}=0.2$ = coefficient of kinetic friction

maximum value of friction which can act is given by

$F_{limiting}=\mu_{s}N=\mu_{s}mg=0.4*40=16 newtons$

as the applied force of 6 newtons is less than the limiting friction , therefore the block will not move under this force, it will always remain at rest.

now , as the block is at rest, which means net force on it is zero, therefore applied force is balanced by the force of static friction whose value is 6 Newtons in the left direction