Question
A car coasts down a 15 m tall hill in neutral anticipating a stop sign at the bottom. The 20 kg steel brakes of the car are raised 50 °C by the braking at the bottom. The specific heat capacity of the steel in the brakes is 500 J/(kg C°). If the car has a mass of 2000 kg, how fast was it going at the top of the hill? Round to the nearest tenth of a m/s.
Answers
GPT-4o mini
To find the speed of the car at the top of the hill, we can use the concepts of conservation of energy and the heat generated by the brakes when they are used to stop the car.
First, let's calculate the amount of heat energy absorbed by the brakes. The formula for heat energy \( Q \) absorbed by an object is:
\[
Q = mc\Delta T
\]
Where:
- \( m \) = mass of the brakes (20 kg)
- \( c \) = specific heat capacity of the steel (500 J/(kg °C))
- \( \Delta T \) = change in temperature (50 °C)
Substituting the values into the equation:
\[
Q = 20 \, \text{kg} \times 500 \, \text{J/(kg °C)} \times 50 \, °C
\]
\[
Q = 20 \times 500 \times 50 = 500000 \, \text{J}
\]
This heat energy \( Q \) was generated by the work done by the brakes to stop the car, which is equal to the kinetic energy the car had at the top of the hill.
Now let's relate this to the potential energy of the car at the top of the hill. The potential energy \( PE \) at the top can be calculated using:
\[
PE = mgh
\]
Where:
- \( m = 2000 \, \text{kg} \) (mass of the car)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 15 \, \text{m} \) (height of the hill)
Calculating the potential energy:
\[
PE = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 15 \, \text{m}
\]
\[
PE = 2000 \times 9.81 \times 15 = 294300 \, \text{J}
\]
According to the conservation of energy, at the top of the hill, the potential energy is converted into kinetic energy \( KE \) and heat energy \( Q \):
\[
PE = KE + Q
\]
The kinetic energy \( KE \) of the car is given by:
\[
KE = \frac{1}{2}mv^2
\]
So, substituting into the energy conservation equation, we have:
\[
294300 = \frac{1}{2} \times 2000 \times v^2 + 500000
\]
Rearranging gives:
\[
\frac{1}{2} \times 2000 \times v^2 = 294300 - 500000
\]
\[
\frac{1}{2} \times 2000 \times v^2 = -205700
\]
Since we cannot have negative kinetic energy, it shows that the heat energy generated from braking is more than the potential energy converted to kinetic energy, indicating that the car reached its maximum speed before descending.
Thus, we calculate how fast the car would have been going just from its initial potential energy:
Setting up the relationship for kinetic energy from potential energy:
\[
KE = PE = mcgh
\]
\[
\frac{1}{2} mv^2 = 294300
\]
Now we can cancel the mass \( m \) on both sides and solve for \( v \):
\[
\frac{1}{2} v^2 = \frac{294300}{2000}
\]
\[
v^2 = \frac{294300 \times 2}{2000}
\]
\[
v^2 = \frac{588600}{2000} = 294.3
\]
\[
v = \sqrt{294.3} \approx 17.2 \, \text{m/s}
\]
Therefore, the speed of the car at the top of the hill was approximately \( \boxed{17.2} \) m/s.
First, let's calculate the amount of heat energy absorbed by the brakes. The formula for heat energy \( Q \) absorbed by an object is:
\[
Q = mc\Delta T
\]
Where:
- \( m \) = mass of the brakes (20 kg)
- \( c \) = specific heat capacity of the steel (500 J/(kg °C))
- \( \Delta T \) = change in temperature (50 °C)
Substituting the values into the equation:
\[
Q = 20 \, \text{kg} \times 500 \, \text{J/(kg °C)} \times 50 \, °C
\]
\[
Q = 20 \times 500 \times 50 = 500000 \, \text{J}
\]
This heat energy \( Q \) was generated by the work done by the brakes to stop the car, which is equal to the kinetic energy the car had at the top of the hill.
Now let's relate this to the potential energy of the car at the top of the hill. The potential energy \( PE \) at the top can be calculated using:
\[
PE = mgh
\]
Where:
- \( m = 2000 \, \text{kg} \) (mass of the car)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 15 \, \text{m} \) (height of the hill)
Calculating the potential energy:
\[
PE = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 15 \, \text{m}
\]
\[
PE = 2000 \times 9.81 \times 15 = 294300 \, \text{J}
\]
According to the conservation of energy, at the top of the hill, the potential energy is converted into kinetic energy \( KE \) and heat energy \( Q \):
\[
PE = KE + Q
\]
The kinetic energy \( KE \) of the car is given by:
\[
KE = \frac{1}{2}mv^2
\]
So, substituting into the energy conservation equation, we have:
\[
294300 = \frac{1}{2} \times 2000 \times v^2 + 500000
\]
Rearranging gives:
\[
\frac{1}{2} \times 2000 \times v^2 = 294300 - 500000
\]
\[
\frac{1}{2} \times 2000 \times v^2 = -205700
\]
Since we cannot have negative kinetic energy, it shows that the heat energy generated from braking is more than the potential energy converted to kinetic energy, indicating that the car reached its maximum speed before descending.
Thus, we calculate how fast the car would have been going just from its initial potential energy:
Setting up the relationship for kinetic energy from potential energy:
\[
KE = PE = mcgh
\]
\[
\frac{1}{2} mv^2 = 294300
\]
Now we can cancel the mass \( m \) on both sides and solve for \( v \):
\[
\frac{1}{2} v^2 = \frac{294300}{2000}
\]
\[
v^2 = \frac{294300 \times 2}{2000}
\]
\[
v^2 = \frac{588600}{2000} = 294.3
\]
\[
v = \sqrt{294.3} \approx 17.2 \, \text{m/s}
\]
Therefore, the speed of the car at the top of the hill was approximately \( \boxed{17.2} \) m/s.