What is the skateboarders mass? **Problem 17: Skateboarder’s Speed and Mass Calculation**A skateboarder reached a speed of 35.0 m/s on his skateboard. Suppose it took 21,500 J of work for him to reach this speed from a speed of 25.0 m/s. Calculate the skateboarder’s mass.**Solution Explanation:**To calculate the skateboarder's mass, we use the work-energy principle. The work done on the skateboarder is equal to the change in kinetic energy:\[ \text{Work} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]Kinetic energy (\( KE \)) is given by the formula:\[ KE = \frac{1}{2}mv^2 \]**Given:**- Initial speed (\( v_i \)) = 25.0 m/s- Final speed (\( v_f \)) = 35.0 m/s- Work done (\( W \)) = 21,500 J**Steps:**1. Calculate the initial kinetic energy (\( KE_{\text{initial}} \)): \[ KE_{\text{initial}} = \frac{1}{2} m (v_i)^2 = \frac{1}{2} m (25.0)^2 \]2. Calculate the final kinetic energy (\( KE_{\text{final}} \)): \[ KE_{\text{final}} = \frac{1}{2} m (v_f)^2 = \frac{1}{2} m (35.0)^2 \]3. Apply the work-energy principle: \[ 21,500 = KE_{\text{final}} - KE_{\text{initial}} \]4. Substitute the kinetic energy expressions and solve for \( m \): \[ 21,500 = \frac{1}{2} m (35.0)^2 - \frac{1}{2} m (25.0)^2 \]5. Simplify and solve for \( m \): \[ 21,500 = \frac{1}{2} m (35.0^2 - 25.0^2) \] \[ 21,500 = \frac{1}{2} m (1225 - 625) \] \[ 21,500 = \frac{1}{2} m (600) \

Name: What is the skateboarders mass? **Problem 17: Skateboarder’s Speed and
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: What is the skateboarders mass? **Problem 17: Skateboarder’s Speed and Mass Calculation**A skateboarder reached a speed of 35.0 m/s on his skateboard. Suppose it took 21,500 J of work for him to reach this speed from a speed of 25.0 m/s. Calculate th