A hockey puck is shot with an initial velocity of 8.20 m/s towards a 52.0° inclined plane 12.0 m away.

A. How high up the incline does the hockey puck slide before coming to rest if the coefficient of friction is 0.250?

B. What is the acceleration of the puck on the way back down the ramp?



Answer :

Let's break down the problem step-by-step:

### Part A: How high up the incline does the hockey puck slide before coming to rest?

1. Given Data:
- Initial velocity ([tex]\(v_0\)[/tex]) = 8.20 m/s
- Incline angle ([tex]\(\theta\)[/tex]) = 52.0°
- Distance along the horizontal to the incline ([tex]\(d\)[/tex]) = 12.0 m
- Coefficient of friction ([tex]\(\mu\)[/tex]) = 0.250
- Acceleration due to gravity ([tex]\(g\)[/tex]) = 9.81 m/s²

2. First, convert the incline angle from degrees to radians:
[tex]\[\theta_\text{rad} = \text{angle in degrees} \times \frac{\pi}{180} = 52.0^\circ \times \frac{\pi}{180}\][/tex]

3. Calculate the components of the initial velocity:
- Parallel to the incline ([tex]\(v_{0\parallel}\)[/tex]):
[tex]\[v_{0\parallel} = v_0 \cos(\theta_\text{rad}) = 8.20 \cos(52.0^\circ)\][/tex]
- Perpendicular to the incline ([tex]\(v_{0\perp}\)[/tex]):
[tex]\[v_{0\perp} = v_0 \sin(\theta_\text{rad}) = 8.20 \sin(52.0^\circ)\][/tex]

4. Calculate the acceleration components along the incline due to gravity and friction:
- Acceleration due to gravity along the incline:
[tex]\[a_g = g \sin(\theta_\text{rad}) = 9.81 \sin(52.0^\circ)\][/tex]
- Acceleration due to friction along the incline:
[tex]\[a_f = g \cos(\theta_\text{rad}) \times \mu = 9.81 \cos(52.0^\circ) \times 0.250\][/tex]

5. Total acceleration along the incline ([tex]\(a_{\text{total}}\)[/tex]):
[tex]\[a_{\text{total}} = a_g + a_f = 9.81 \sin(52.0^\circ) + 9.81 \cos(52.0^\circ) \times 0.250\][/tex]

6. Determine the distance traveled up the incline ([tex]\(d_\text{up}\)[/tex]):
Using the kinematic equation [tex]\(v_f^2 = v_i^2 + 2a s\)[/tex], where [tex]\(v_f\)[/tex] (final velocity) is 0 when the puck comes to rest:
[tex]\[0 = v_{0\parallel}^2 - 2 \times a_{\text{total}} \times d_\text{up}\][/tex]
Solving for [tex]\(d_\text{up}\)[/tex]:
[tex]\[d_\text{up} = \frac{v_{0\parallel}^2}{2 \times a_{\text{total}}}\][/tex]

Plugging in the previously calculated values:
[tex]\[d_\text{up} = \frac{5.048424097670398^2}{2 \times 9.24029526111812}\][/tex]
[tex]\[d_\text{up} \approx 1.3791001883448024 \, \text{m}\][/tex]

### Part B: What is the acceleration of the puck on the way back down the ramp?

1. Calculate the net acceleration on the way back down:
On the way down, gravity aids the motion while friction opposes it. Thus, the net acceleration ([tex]\(a_{\text{down}}\)[/tex]) is:
[tex]\[a_{\text{down}} = g \sin(\theta_\text{rad}) - g \cos(\theta_\text{rad}) \times \mu\][/tex]

2. Plug in the values:
[tex]\[a_{\text{down}} = 9.81 \sin(52.0^\circ) - 9.81 \cos(52.0^\circ) \times 0.250\][/tex]
[tex]\[a_{\text{down}} \approx 6.220475724645766 \, \text{m/s}^2\][/tex]

In summary:
- (A) The hockey puck slides approximately 1.379 m up the incline before coming to rest.
- (B) The acceleration of the puck on the way back down the ramp is approximately 6.220 m/s².