Friction Force — Pushing a Crate at Constant Speed (Worked Example)

Easy

Problem

A warehouse worker pushes a 40 kg crate across a level concrete floor at constant velocity. The coefficient of kinetic friction between crate and floor is 0.35. What horizontal push force is required?

Free-body diagram of the crate sliding at constant velocity.

Given

m = 40 kg
μk = 0.35
velocity is constant (a = 0)
g = 9.81 m/s²

Find

The kinetic friction force and the required push.

Principle

Kinetic friction is f_k = μk·N. At constant velocity the net force is zero, so the push must exactly balance friction.

Formula

f_k = \mu_k N, \qquad N = mg

f_k = μ_k N, with N = mg on level ground

Solution

Normal force: $N = mg = 40 \times 9.81 = 392.4\ \mathrm{N}$
Friction: $f_k = \mu_k N = 0.35 \times 392.4$
Calculate: $f_k = 137.34\ \mathrm{N} \approx 137\ \mathrm{N}$
Constant velocity means $F_{net} = 0$ , so the push $= 137\ \mathrm{N}$ forward

≈ 1.37 × 10² N (≈ 137 N) of push

Unit check

μk is unitless; N is in newtons, so μk·N is in newtons. Matches force.

What this means physically

About 137 N — roughly the weight of a 14 kg object — must be pushed continuously just to fight friction. Heavier crates or rougher floors raise this directly.

Common mistake

Using the static friction coefficient for a moving crate, or forgetting that at constant velocity the push equals friction (not mass times some acceleration).