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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
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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