Question25. (II) A package of mass $m$ is dropped vertically onto a horizontal conveyor belt whose speed is $v=1.5 \mathrm{~m} / \mathrm{s}$, and the coefficient of kinetic friction between the package and the belt is $\mu_{\mathrm{k}}=0.70$. (a) For how much time does the package slide on the belt (until it is at rest relative to the belt)? (b) How far does the package move during this time?

Studdy Solution
Determine the distance the package moves during this time.
Using the equation for distance $s = ut + \frac{1}{2}at^2$, where $u = 0 \, \text{m/s}$: $$s = 0 \cdot t + \frac{1}{2} \cdot 6.86 \cdot (0.219)^2$$
Calculate $s$: $$s = \frac{1}{2} \cdot 6.86 \cdot 0.048 \approx 0.165 \, \text{m}$$
The package slides on the belt for approximately $0.219 \, \text{s}$ and moves a distance of approximately $0.165 \, \text{m}$.