Question
A bobsled team gets a sled going 6 m/s at the top of a bobsled run that is 130 m high. The sled and crew has a mass of 1600 kg. At the bottom of the run the brakeman on the team deploys the sled’s brakes which increase the negligible friction on the of the sled on the ice. The sled needs 300 m to come to a stop, What is effective coefficient of friction of the brake with the ice? Select the kinds of energy that you can use in this problem. Select all that apply.
Responses
Work
Work
Thermal Energy
Thermal Energy
Kinetic Energy
Kinetic Energy
Gravity Potential Energy
Gravity Potential Energy
Spring Potential Energy
Responses
Work
Work
Thermal Energy
Thermal Energy
Kinetic Energy
Kinetic Energy
Gravity Potential Energy
Gravity Potential Energy
Spring Potential Energy
Answers
Answered by
GPT-4o mini
In this scenario, we can analyze the energy transformations and calculate the effective coefficient of friction. The energies that apply in this situation are:
1. **Kinetic Energy**: The sled has a certain speed, which gives it kinetic energy at the top of the run.
2. **Gravity Potential Energy**: The sled starts at a height of 130 m, giving it gravitational potential energy.
3. **Thermal Energy**: When the brakes are applied, some of the kinetic energy is converted into thermal energy due to friction.
The spring potential energy is not applicable as we are not dealing with a spring system in this problem.
Based on this, the selected energies that apply are:
- Kinetic Energy
- Gravity Potential Energy
- Thermal Energy
Now to calculate the effective coefficient of friction, we will use the kinetic energy just before the brakes are applied, and the work done by friction to bring the sled to a stop.
1. Calculate the initial kinetic energy (KE) at the bottom of the run.
The kinetic energy is given by:
\[
KE = \frac{1}{2} mv^2
\]
where \( m = 1600 \, \text{kg} \) and \( v = 6 \, \text{m/s} \).
2. Calculate the work done by friction (W) while stopping the sled:
\[
W = f_{\text{friction}} \times d
\]
where \( d = 300 \, \text{m} \) and \( f_{\text{friction}} = \mu m g \), with \( g \approx 9.81 \, \text{m/s}^2 \).
3. Set the initial kinetic energy equal to the work done by friction:
\[
\frac{1}{2} mv^2 = \mu m g d
\]
We can simplify this equation (mass \( m \) cancels out):
\[
\frac{1}{2} v^2 = \mu g d
\]
\[
\mu = \frac{\frac{1}{2} v^2}{g d}
\]
Now, using the values:
- \( v = 6 \, \text{m/s} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( d = 300 \, \text{m} \)
Substituting the values:
\[
\mu = \frac{\frac{1}{2} (6)^2}{9.81 \times 300}
\]
\[
\mu = \frac{18}{2943}
\]
\[
\mu \approx 0.0061
\]
Thus, the effective coefficient of friction of the brake with the ice is approximately **0.0061**.
1. **Kinetic Energy**: The sled has a certain speed, which gives it kinetic energy at the top of the run.
2. **Gravity Potential Energy**: The sled starts at a height of 130 m, giving it gravitational potential energy.
3. **Thermal Energy**: When the brakes are applied, some of the kinetic energy is converted into thermal energy due to friction.
The spring potential energy is not applicable as we are not dealing with a spring system in this problem.
Based on this, the selected energies that apply are:
- Kinetic Energy
- Gravity Potential Energy
- Thermal Energy
Now to calculate the effective coefficient of friction, we will use the kinetic energy just before the brakes are applied, and the work done by friction to bring the sled to a stop.
1. Calculate the initial kinetic energy (KE) at the bottom of the run.
The kinetic energy is given by:
\[
KE = \frac{1}{2} mv^2
\]
where \( m = 1600 \, \text{kg} \) and \( v = 6 \, \text{m/s} \).
2. Calculate the work done by friction (W) while stopping the sled:
\[
W = f_{\text{friction}} \times d
\]
where \( d = 300 \, \text{m} \) and \( f_{\text{friction}} = \mu m g \), with \( g \approx 9.81 \, \text{m/s}^2 \).
3. Set the initial kinetic energy equal to the work done by friction:
\[
\frac{1}{2} mv^2 = \mu m g d
\]
We can simplify this equation (mass \( m \) cancels out):
\[
\frac{1}{2} v^2 = \mu g d
\]
\[
\mu = \frac{\frac{1}{2} v^2}{g d}
\]
Now, using the values:
- \( v = 6 \, \text{m/s} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( d = 300 \, \text{m} \)
Substituting the values:
\[
\mu = \frac{\frac{1}{2} (6)^2}{9.81 \times 300}
\]
\[
\mu = \frac{18}{2943}
\]
\[
\mu \approx 0.0061
\]
Thus, the effective coefficient of friction of the brake with the ice is approximately **0.0061**.