Question

A 20 kg box moving at an initial speed of 10 m/s slides 25 m to the right on a horizontal floor before it comes to a complete stop. What is the coefficient of friction between the box and the floor?

Answer 1

The coefficient of friction between the 20 kg box and the floor is 0.2

We'll begin by calculating the acceleration of the box. This can be obtained as follow:

Initial velocity (u) = 10 m/s

Final velocity (v) = 0 m/s

Distance (s) = 25 m

Acceleration (a) =?

v² = u² + 2as

0² = 10² + (2 × a × 25)

0 = 100 + 50a

Collect like terms

0 – 100 = 50a

–100 = 50a

Divide both side by 50

a = –2 m/s²

Finally, we shall determine the coefficient of friction. This can be obtained as follow:

Mass (m) = 20 Kg

Acceleration (a) = –2 m/s²

Acceleration due to gravity (g) = 10 m/s²

Frictional Force (F) = –ma = – (20 × –2) = 40 N

Normal reaction (N) = mg = 20 × 10 = 200 N

Coefficient of friction (μ) =?

$\mu = \frac{F}{N} \\\\\mu = \frac{40}{200}$

μ = 0.2

Therefore, the coefficient of friction between the box and the floor is 0.2

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Answer 2

The coefficient of friction between the box and the floor is 0.204.

Given data:

The mass of box is, m = 20 kg.

The initial speed of box is, u = 10 m/s.

The sliding distance is, s = 25 m.

The linear force acting on the fox is providing it necessary friction. Then,

$F = -F_{f}\\ma = -\mu mg\\a= -\mu \times g$

Here, a is the linear acceleration and $\mu$ is the coefficient of friction between the box and the floor.

Applying the third kinematic equation of motion as,

$v^{2}=u^{2}+2as\\0=10^{2}+2(a)\times 25\\a = \dfrac{-100}{50}\\a =- 2\;\rm m/s^{2}$

Then,

$a=- \mu \times g\\-2= -\mu \times 9.8\\\mu = 0.204$

Thus, we can conclude that the coefficient of friction between the box and the floor is 0.204.

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