A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N of friction. What distance does the box slide?

the distance moved by the box is 70.03 m.

Explanation:

Given;

mass of the box, m = 35 kg

initial velocity of the box, u = 10 m/s

frictional force, F = 25 N

Apply Newton's second law of motion to determine the deceleration of the box;

-F = ma

a = -F / m

a = (-25 ) / 35

a = -0.714 m/s²

The distance moved by the box is calculated as follows;

v² = u² + 2ad

where;

v is the final velocity of the box when it comes to rest = 0

0 = 10² + (2 x - 0.714)d

0 = 100 - 1.428d

1.428d = 100

d = 100 / 1.428

d = 70.03 m

Therefore, the distance moved by the box is 70.03 m.

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