Questions: Jim rides his skateboard down a ramp that is in the shape of a quarter circle with a radius of 5.00 m. At the bottom of the ramp, Jim is moving at 9.00 m/s. Jim and his skateboard have a mass of 72.1 kg. How much work is done by friction as the skateboard goes down the ramp?

Solution Steps

The potential energy (PE) at the top of the ramp is given by the formula: \[ PE = mgh \] where \( m = 72.1 \, \text{kg} \) is the mass, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h \) is the height of the ramp. Since the ramp is a quarter circle with radius \( r = 5.00 \, \text{m} \), the height \( h \) is equal to the radius: \[ h = 5.00 \, \text{m} \] Thus, the potential energy is: \[ PE = 72.1 \times 9.81 \times 5.00 = 3535.305 \, \text{J} \]

The kinetic energy (KE) at the bottom of the ramp is given by the formula: \[ KE = \frac{1}{2}mv^2 \] where \( v = 9.00 \, \text{m/s} \) is the velocity at the bottom. Thus, the kinetic energy is: \[ KE = \frac{1}{2} \times 72.1 \times (9.00)^2 = 2916.45 \, \text{J} \]

The work done by friction (W_f) is the difference between the initial potential energy and the final kinetic energy: \[ W_f = PE - KE \] Substituting the values we calculated: \[ W_f = 3535.305 - 2916.45 = 618.855 \, \text{J} \]

Final Answer