A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable that is parallel to the ground. The cable pulls her a distance of 60.0 m along a 30.0° frictionless slope at a constant speed of 2.00 m/s. (a) How much time does it take to get up the slope? (b) What is her change in Gravitational Potential Energy?
A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable that is parallel to the ground. The cable pulls her a distance of 60.0 m along a 30.0° frictionless slope at a constant speed of 2.00 m/s. (a) How much time does it take to get up the slope? (b) What is her change in Gravitational Potential Energy?
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![### Problem Statement:
**10.** A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable that is parallel to the ground. The cable pulls her a distance of 60.0 m along a 30.0° frictionless slope at a constant speed of 2.00 m/s.
**(a)** How much time does it take to get up the slope?
**(b)** What is her change in Gravitational Potential Energy?
### Solution Explanation:
#### **(a) Time to get up the slope:**
To find the time it takes for the skier to get up the slope, we need to use the formula:
\[ t = \frac{d}{v} \]
Where:
- \( d \) is the distance pulled along the slope, which is 60.0 meters.
- \( v \) is the constant speed, which is 2.00 meters per second (m/s).
Substituting the given values:
\[ t = \frac{60.0 \, \text{m}}{2.00 \, \text{m/s}} = 30.0 \, \text{seconds} \]
#### **(b) Change in Gravitational Potential Energy (GPE):**
To calculate the change in gravitational potential energy, we use the equation:
\[ \Delta \text{GPE} = mgh \]
Where:
- \( m \) is the mass of the skier, which is 70.0 kg.
- \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
- \( h \) is the change in height.
The change in height (\( h \)) can be determined using the sine component of the slope angle (\( \theta \)):
\[ h = d \sin(\theta) \]
Where:
- \( d \) is the distance along the slope (60.0 m).
- \( \theta \) is the angle of the slope (30.0°).
Substituting in these values:
\[ h = 60.0 \, \text{m} \times \sin(30.0°) = 60.0 \, \text{m} \times 0.5 = 30.0 \, \text{m} \]
Now we can calculate the change in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1829f362-1a16-45e8-8282-990908270dc7%2F7174e443-905e-4fb9-96f0-06bbc5e0ad60%2Fgzqisc7_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**10.** A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable that is parallel to the ground. The cable pulls her a distance of 60.0 m along a 30.0° frictionless slope at a constant speed of 2.00 m/s.
**(a)** How much time does it take to get up the slope?
**(b)** What is her change in Gravitational Potential Energy?
### Solution Explanation:
#### **(a) Time to get up the slope:**
To find the time it takes for the skier to get up the slope, we need to use the formula:
\[ t = \frac{d}{v} \]
Where:
- \( d \) is the distance pulled along the slope, which is 60.0 meters.
- \( v \) is the constant speed, which is 2.00 meters per second (m/s).
Substituting the given values:
\[ t = \frac{60.0 \, \text{m}}{2.00 \, \text{m/s}} = 30.0 \, \text{seconds} \]
#### **(b) Change in Gravitational Potential Energy (GPE):**
To calculate the change in gravitational potential energy, we use the equation:
\[ \Delta \text{GPE} = mgh \]
Where:
- \( m \) is the mass of the skier, which is 70.0 kg.
- \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
- \( h \) is the change in height.
The change in height (\( h \)) can be determined using the sine component of the slope angle (\( \theta \)):
\[ h = d \sin(\theta) \]
Where:
- \( d \) is the distance along the slope (60.0 m).
- \( \theta \) is the angle of the slope (30.0°).
Substituting in these values:
\[ h = 60.0 \, \text{m} \times \sin(30.0°) = 60.0 \, \text{m} \times 0.5 = 30.0 \, \text{m} \]
Now we can calculate the change in
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