Problem — Kinetic Friction from the a–F Graph

Problem

A crate of mass $m = 5{,}0\;\text{kg}$ is pulled horizontally across a floor. The graph in the figure shows the measured acceleration as the applied force varies. Interpret the graph and derive the coefficient of kinetic friction (use $g = 9{,}8\;\text{m/s}^2$).

Solution

Step 1 — Interpreting the graph. As long as the applied force is small, the crate stays at rest ($a = 0$): static friction holds it back. Only above a threshold value does the crate start to move, and the acceleration grows linearly with the force. The line indeed starts from $a = 0$ at the threshold, which from the graph reads

$F_{\text{threshold}} \approx 10\;\text{N}$

Step 2 — Model. Once moving, the applied force $F$ and the kinetic friction $f_d$, constant, act on the crate. Newton’s second law along the floor gives

$F - f_d = m\,a \quad\Longrightarrow\quad a = \frac{F - f_d}{m}$

This is exactly a straight line in the $a$–$F$ plane: it vanishes ($a = 0$) when $F = f_d$. The threshold value read off the graph therefore coincides with the kinetic friction force:

$f_d \approx 10\;\text{N}$

Step 3 — Coefficient of kinetic friction. On a horizontal plane the normal reaction is $R = mg$, so

$\mu_d = \frac{f_d}{R} = \frac{f_d}{m\,g} = \frac{10\;\text{N}}{5{,}0\;\text{kg} \times 9{,}8\;\text{m/s}^2} = \frac{10}{49} \approx 0{,}20$

$\ev{\mu_d \approx 0{,}20}$

Links

Topics: Friction Concepts: Kinetic friction Skills: Reading graphs