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

STEP 1

1. The package is initially at rest relative to the ground when it is dropped onto the conveyor belt.
2. The conveyor belt moves at a constant speed of $v = 1.5 \, \text{m/s}$ .
3. The coefficient of kinetic friction between the package and the belt is $\mu_k = 0.70$ .
4. The package slides on the belt until it reaches the same speed as the belt, at which point it is at rest relative to the belt.
5. We are to find the time the package slides on the belt and the distance it moves during this time.

STEP 2

1. Determine the acceleration of the package due to friction.
2. Calculate the time it takes for the package to match the speed of the belt.
3. Determine the distance the package moves during this time.

STEP 3

Determine the acceleration of the package due to friction.
The force of kinetic friction $f_k$ acting on the package is given by: $f_k = \mu_k \cdot m \cdot g$
The acceleration $a$ of the package is given by Newton's second law: $a = \frac{f_k}{m} = \frac{\mu_k \cdot m \cdot g}{m} = \mu_k \cdot g$
Substitute the given values: $a = 0.70 \times 9.8 \, \text{m/s}^2 = 6.86 \, \text{m/s}^2$

STEP 4

Calculate the time it takes for the package to match the speed of the belt.
The initial velocity of the package relative to the belt is $0 \, \text{m/s}$ , and it needs to reach $v = 1.5 \, \text{m/s}$ .
Using the equation $v = u + at$ , where $u = 0 \, \text{m/s}$ , we have: $1.5 = 0 + 6.86 \cdot t$
Solve for $t$ : $t = \frac{1.5}{6.86} \approx 0.219 \, \text{s}$

STEP 5

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 $0.165 \, \text{m}$ .

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