Questions: A 60.8-kg skateboarder starts out with a speed of 2.49 m/s. He does 101 J of work on himself by pushing with his feet against the ground. In addition, friction does -246 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 9.34 m/s. (a) Calculate the change (PEf-PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.

Solution Steps

We need to calculate the change in gravitational potential energy and the change in vertical height of a skateboarder. The skateboarder has an initial speed, does work on himself, and experiences work done by friction. We are given the initial and final speeds, the work done by the skateboarder, and the work done by friction.

The work-energy principle states that the change in kinetic energy plus the change in potential energy is equal to the total work done on the system. Mathematically, this is expressed as:

\[ \Delta KE + \Delta PE = W_{\text{total}} \]

Where:

• \(\Delta KE\) is the change in kinetic energy.
• \(\Delta PE\) is the change in potential energy.
• \(W_{\text{total}}\) is the total work done on the skateboarder.

The change in kinetic energy (\(\Delta KE\)) is given by:

\[ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_0^2 \]

Substitute the given values:

• \(m = 60.8 \, \text{kg}\)
• \(v_0 = 2.49 \, \text{m/s}\)
• \(v_f = 9.34 \, \text{m/s}\)

\[ \Delta KE = \frac{1}{2} \times 60.8 \times (9.34^2) - \frac{1}{2} \times 60.8 \times (2.49^2) \]

\[ \Delta KE = 2670.5088 - 188.3808 = 2482.1280 \, \text{J} \]

The total work done (\(W_{\text{total}}\)) is the sum of the work done by the skateboarder and the work done by friction:

\[ W_{\text{total}} = 101 \, \text{J} - 246 \, \text{J} = -145 \, \text{J} \]

Using the work-energy principle:

\[ \Delta KE + \Delta PE = W_{\text{total}} \]

\[ 2482.1280 + \Delta PE = -145 \]

\[ \Delta PE = -145 - 2482.1280 = -2627.1280 \, \text{J} \]

The change in potential energy is related to the change in height (\(\Delta h\)) by:

\[ \Delta PE = mgh \]

Solving for \(\Delta h\):

\[ \Delta h = \frac{\Delta PE}{mg} = \frac{-2627.1280}{60.8 \times 9.81} \]

\[ \Delta h = \frac{-2627.1280}{596.448} = -4.4030 \, \text{m} \]

The absolute value of the change in height is \(4.4030 \, \text{m}\).

Final Answer