### 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

(a) How much time does it take to get up the slope. (b) What is her change in Gravitational…

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