Question

A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Once the crate is in motion, a horizontal force of 53 N keeps the crate moving with a constant velocity. Find μk, the coefficient of kinetic friction, between the crate and the floor.

Answer 1

Answer:

The coefficient of kinetic friction is approximately 0.225

Explanation:

The given parameters are;

The mass of the crate = 24 kg

The force required to set the mass in motion = 75 N

The force required to keep the crate in constant velocity = 53 N

We have;

Sum of horizontal forces, ∑Fₓ = 0

∑Fₓ = The force pulling thing the crate - The force of friction - The force of acceleration of the crate

The force of acceleration of the crate = 0 (Crate with constant velocity)

The force of friction = $\mu_k$ × N

N = The normal (acting perpendicular to the surface) force = (Here) Mass of crate × Gravity

$\mu_k$ = The coefficient of kinetic friction

N = 24 × 9.81 = 235.44 N

The force of friction = $\mu_k$ × 235.44

∴ ∑Fₓ = The force pulling thing the crate - The force of friction

∑Fₓ = 53 - $\mu_k$ × 235.44 = 0

53 - $\mu_k$ × 235.44 = 0

53 = $\mu_k$ × 235.44

$\mu_k$ = 53/235.44 ≈ 0.225

$\mu_k$ = The coefficient of kinetic friction ≈ 0.225.

Answer 2

The coefficient of kinetic friction between the crate and the floor is 0.22.

The coefficient of kinetic friction μk shows the ratio between frictional force and normal force over a body on a rough surface.

Given that:

• mass of the crate = 24 kg

• the horizontal force $\mathbf{F_{s,max}}$ = 75 N

We need to first find the coefficient of static friction (μs) to determine (μk);

Using the formula:

$\mathbf{\mu_s = \dfrac{F_{s, max}}{m\times g}}$

$\mathbf{\mu_s = \dfrac{75 \ N }{24 \ kg \times 9.81 \ m/s^2}}$

$\mathbf{\mu_s = \dfrac{75 }{235.44}}$

$\mathbf{\mu_s =0.32 }$

• Now, once the crate is in motion; the horizontal force = 53 N
• However, the net force  = 0; since F = Fk

∴

The coefficient of kinetic friction can be computed as;

$\mathbf{F = \mu_k \times N}$

$\mathbf{ \mu_k = \dfrac{F}{N}}$

$\mathbf{ \mu_k = \dfrac{f}{m \times g}}$

$\mathbf{ \mu_k = \dfrac{53}{24 \times 9.8}}$

$\mathbf{ \mu_k =0.22 }$

Therefore, we can conclude that the coefficient of the kinetic friction is 0.22

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