A skier (mass = 65 kg) is at rest at the top of a 75 m hill. She pushes off and when she reaches the bottom of the hill she has a speed of 25 m/s.

Question

A skier (mass = 65 kg) is at rest at the top of a 75 m hill. She pushes off and when she reaches the bottom of the hill she has a speed of 25 m/s. 
(a) What type of energy transformation has taken place?

(b) Find the efficiency of the energy transformation.

GPT-4o mini · Answer

To analyze the skier's situation, we first need to understand the energy transformations that occur and then calculate the efficiency of this process.

(a) Type of Energy Transformation:

Initially, the skier has gravitational potential energy (PE) at the top of the hill due to her height. As she descends, this potential energy is converted into kinetic energy (KE). The transformation process can be summarized as follows:

At the top: The skier has gravitational potential energy given by the formula: \[ PE = mgh \] where:
- \( m = 65 , \text{kg} \) (mass of the skier)
- \( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 75 , \text{m} \) (height of the hill)
At the bottom: The skier has kinetic energy: \[ KE = \frac{1}{2} mv^2 \] where \( v = 25 , \text{m/s} \) (final speed of the skier).

Thus, the energy transformation process is from gravitational potential energy to kinetic energy.

(b) Efficiency of the Energy Transformation:

Step 1: Calculate the initial gravitational potential energy (PE). \[ PE = mgh = 65 , \text{kg} \times 9.81 , \text{m/s}^2 \times 75 , \text{m} \] \[ PE = 65 \times 9.81 \times 75 \approx 48037.5 , \text{J} , \text{(joules)} \]

Step 2: Calculate the final kinetic energy (KE) at the bottom. \[ KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 65 , \text{kg} \times (25 , \text{m/s})^2 \] \[ KE = \frac{1}{2} \times 65 \times 625 = 20312.5 , \text{J} \]

Step 3: Calculate the efficiency. The efficiency (\( \eta \)) can be calculated using: \[ \eta = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100% \] In this case, the useful energy output is the kinetic energy at the bottom, and the total energy input is the gravitational potential energy at the top: \[ \eta = \frac{KE}{PE} \times 100% = \frac{20312.5 , \text{J}}{48037.5 , \text{J}} \times 100% \] \[ \eta \approx 42.35% \]

Summary of Answers:

(a) The type of energy transformation is from gravitational potential energy to kinetic energy.

(b) The efficiency of the energy transformation is approximately 42.35%.