Friction — Question 10

Question 10

Relative motion can stop even while both objects accelerate

A 1 kg parcel is placed with ground speed zero on a belt initially moving right at 2 m/s. The belt accelerates right at 1 m/s$^2$. Kinetic friction on the parcel is 3 N and its static limit is 2 N. Treat the belt motion as imposed.

1. Find parcel and belt velocities while slipping.

2. Find the catch-up time in velocity and the accumulated relative sliding distance.

3. Can the parcel stay with the belt afterward, and what is its friction then?

Question 10 - Solution

Relative motion can stop even while both objects accelerate

(a) Use different accelerations.

Before velocities match, rightward kinetic friction gives $a_p=3/1=3$ m/s$^2$. Thus, with $t$ in seconds, $$\boxed{v_p(t)=3t,\qquad v_b(t)=2+t}\quad\mathrm{m/s}.$$

(b) Close the relative-speed gap.

Equate the two velocities: $$3t_c=2+t_c\Longrightarrow\boxed{t_c=1\ \mathrm s},\qquad v_c=3\ \mathrm{m/s}.$$ The accumulated contact sliding distance is $$\boxed{s_{\mathrm{rel}}=\int_0^1(v_b-v_p)\,dt=\int_0^1(2-2t)\,dt=1\ \mathrm m}.$$

(c) Test the new constraint.

Once the velocities match, sticking requires the belt acceleration: $$\boxed{f_s=ma_b=1(1)=1\ \mathrm N\text{ rightward}}<2\ \mathrm N.$$ The static constraint is feasible and persists under the stated conditions. Matching velocities therefore need not imply zero acceleration or zero friction.