Question

A 5 kg block sits on a rough horizontal surface. a force of magnitude f = 10 n acting parallel to the surface is applied to the block. the coefficient of static and kinetic friction between the block and the surface are μs = 0.5 and μk = 0.4, respectively. what is the magnitude (in n) of the friction force acting on the block

Answer 1

The magnitude of the friction force acting on the block is 24.5 N.

Step-by-step explanation:

The magnitude of the friction force acting on the block can be calculated using the equation:
f = μsN

Given that the coefficient of static friction (μs) is 0.5 and the normal force (N) is equal to the weight of the block, which can be calculated as:
N = mg = 5 kg × 9.8 m/s²

Substituting the values into the equation, we have:
f = 0.5 × (5 kg × 9.8 m/s²)

Therefore, the magnitude of the friction force acting on the block is 24.5 N.