A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3m.

1.8 m/s

Explanation:

Draw a free body diagram of the block. There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn - mg = 0

Fn = mg

Sum the forces in the x direction:

F - Fn μ = ma

F - mg μ = ma

a = (F - mg μ) / m

a = (12 N - 6 kg * 9.8 m/s² * 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s) ² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s