Question

Suppose a 15 N force is applied to the side of a 5.0 kg block that is sitting on a table. The block experiences a frictional force against the force that is applied.

a) What is the weight of the block (FG)?

b) What is the normal force on the block (FN)?

c) If the coefficient of kinetic friction is μk = .10, what is the frictional force on the block (Ff)?

d) What is the net force on the block?

e) What is the acceleration of the block from the net force?

Answer 1

Step-by-step explanation:

Fg = W = mg = (5.0 kg)(9.8 m/s²) = 49 N

Fnormal = W = 49 N

Ff = μFn = (0.10)(49 N) = 4.9 N

Fnet = 15 N - 4.9 N = 10.1 N

a = Fnet/m = (10.1N)/(5.0kg) = 2.02 m/s²