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Sagot :
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.
### 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.
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