A 24-N horizontal force is applied to a 40-N block initially at rest on a rough horizontal surface.If the coefficients of friction are = 0.5 and = 0.4, the magnitude of the frictional force on the block is:
A)12 N
B)16 N
C)20 N
D)24 N
E)40 N

Question

A 24-N horizontal force is applied to a 40-N block initially at rest on a rough horizontal surface.If the coefficients of friction are  = 0.5 and  = 0.4, the magnitude of the frictional force on the block is:
A)12 N 
B)16 N 
C)20 N 
D)24 N 
E)40 N

Quizplus · Accepted Answer

First, we need to calculate the maximum force of static friction (fs) which is given by:
fs = μsN
where μs is the coefficient of static friction and N is the normal force, which is equal to the weight of the block in this case: 
N = mg = 40 N

fs1 = μs1N = 0.5 × 40 N = 20 N
fs2 = μs2N = 0.4 × 40 N = 16 N

Since the applied force of 24 N is greater than the maximum force of static friction in both cases, the block will start to move and kinetic friction will come into play. The force of kinetic friction is given by: 
fk = μkN
where μk is the coefficient of kinetic friction, which is the same as the coefficient of static friction in this case: 
μk = μs

Therefore, the magnitude of the frictional force on the block is equal to the force of kinetic friction: 
fk = μkN = μsN = 0.4 × 40 N = 16 N

Therefore, the correct answer is B.

A 24-N Horizontal Force Is Applied to a 40-N Block