A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the surface which causes the block to come to a complete stop. The coefficient of kinetic friction between the block and the surface is 0.5. Calculate the distance the block slides while coming to a stop. Show all work.

• Net vertical force on the block:

∑ F = n - w = 0

(n = magnitude of normal force, w = weight)

n = w = m g

(m = mass, g = 9.8 m/s²)

n = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ F = -f = m a

(f = mag. of friction, a = acceleration)

We have f = µ n = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) a

a = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆x in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆x

∆x = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

inverse v=i*r

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