Friction — Question 1

Question 1

Friction can accelerate a moving object forward

Take $g=10$ m/s$^2$. A 2 kg parcel slides right at 1 m/s on a conveyor moving right at a constant 4 m/s. Kinetic friction has magnitude 6 N. Once velocities match, the parcel can stick. No other horizontal force acts.

1. Find the direction of friction and the parcel acceleration while it slips.

2. Find the time to match the belt speed and the parcel displacement in that time.

3. Find the sliding distance relative to the belt and the friction force after matching.

Question 1 - Solution

Friction can accelerate a moving object forward

(a) Compare the two velocities.

Take right as positive. The contact slip is $$v_{p/b}=v_p-v_b=1-4=-3\ \mathrm{m/s}.$$ Friction opposes this relative motion, so $f_k=+6$ N. Newton’s second law gives $$\boxed{a_p=\frac{f_k}{m}=\frac62=+3\ \mathrm{m/s^2}}.$$

(b) Follow the ground-frame motion.

While slipping, $v_p(t)=1+3t$ in SI units. End this phase when $v_p=v_b$: $$1+3t_c=4\Longrightarrow\boxed{t_c=1\ \mathrm s}.$$ $$\boxed{\Delta x_p=v_{p0}t_c+\tfrac12a_pt_c^2=1(1)+\tfrac12(3)(1)^2=2.5\ \mathrm m}.$$

(c) Compare travel and then test sticking.

The belt travels $\Delta x_b=4(1)=4$ m. Hence $$\Delta x_{p/b}=2.5-4=-1.5\ \mathrm m,\qquad\boxed{s_{\mathrm{slip}}=1.5\ \mathrm m}.$$ After matching, the common velocity is constant: $a_p=0$. Therefore $\boxed{f_s=ma_p=0}$. Friction opposes contact slip, not necessarily ground-frame velocity.