High School

A horizontal force of 92.2 N is applied to a 46.5 kg crate on a rough, level surface. If the crate accelerates at 1.18 m/s\(^2\), what is the magnitude of the force of kinetic friction (in N) acting on the crate?

Answer :

The magnitude of the force of kinetic friction acting on the crate is approximately 37.33 N.

To find the magnitude of the force of kinetic friction acting on the crate, we can use the formula for kinetic friction:

Force of kinetic friction = μ_k * Normal force

where μ_k is the coefficient of kinetic friction and Normal force is the force exerted by the surface on the crate perpendicular to the surface.

The normal force can be calculated as the weight of the crate since it is on a level surface:

Normal force = mass of crate * acceleration due to gravity

Normal force = 46.5 kg * 9.8 m/s^2

Normal force ≈ 455.7 N

Now, we need to find the coefficient of kinetic friction (μ_k). The coefficient of kinetic friction is the ratio of the force of kinetic friction to the normal force.

μ_k = Force of kinetic friction / Normal force

μ_k = Force of kinetic friction / 455.7 N

Given that the crate is accelerating at 1.18 m/s^2, the net force acting on it can be calculated as:

Net force = mass of crate * acceleration

Net force = 46.5 kg * 1.18 m/s^2

Net force ≈ 54.87 N

Now, we can find the force of kinetic friction:

Force of kinetic friction = Net force - Applied force

Force of kinetic friction = 54.87 N - 92.2 N

Force of kinetic friction ≈ -37.33 N

The negative sign indicates that the force of kinetic friction acts in the opposite direction to the applied force. Taking the magnitude, we get:

Magnitude of force of kinetic friction ≈ |-37.33 N| ≈ 37.33 N

So, the magnitude of the force of kinetic friction acting on the crate is approximately 37.33 N.

Learn more about kinetic friction here:

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