A20 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?

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

v² = u² + 2as

0² = 10² + (2 × a × 25)

0 = 100 + 50a

Collect like terms

0 – 100 = 50a

–100 = 50a

Divide both side by 50

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

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

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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,

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

Applying the third kinematic equation of motion as,

Then,

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

Learn more about frictional force here:

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c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

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