Answer :
To determine the speed of the car just before the brakes were applied, we need to use the principles of energy conversion. Here is a detailed, step-by-step solution:
1. Understanding the Problem:
- We are given the total mass of the car: [tex]\( m = 1600 \)[/tex] kg.
- The heat produced when the car is brought to rest: [tex]\( Q = 500 \)[/tex] kJ.
- The car is brought to rest, which implies that its initial kinetic energy was fully converted into heat energy.
2. Conversion of Units:
- Convert the heat produced from kilojoules (kJ) to joules (J). Since [tex]\( 1 \)[/tex] kJ [tex]\( = 1000 \)[/tex] J, we have:
[tex]\[ Q = 500 \text{ kJ} = 500 \times 1000 \text{ J} = 500{,}000 \text{ J} \][/tex]
3. Using the Work-Energy Principle:
- The kinetic energy ([tex]\( KE \)[/tex]) of the car before the brakes were applied is converted into heat energy.
- The formula for kinetic energy is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
- We know that the kinetic energy is equal to the heat energy produced:
[tex]\[ \frac{1}{2} m v^2 = Q \][/tex]
4. Solving for the Initial Speed [tex]\( v \)[/tex]:
- Plug in the known values (mass [tex]\( m \)[/tex] and heat energy [tex]\( Q \)[/tex]):
[tex]\[ \frac{1}{2} \times 1600 \times v^2 = 500{,}000 \][/tex]
- Simplify and solve for [tex]\( v^2 \)[/tex]:
[tex]\[ 800 \times v^2 = 500{,}000 \][/tex]
[tex]\[ v^2 = \frac{500{,}000}{800} \][/tex]
[tex]\[ v^2 = 625 \][/tex]
- Taking the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{625} \][/tex]
[tex]\[ v = 25 \text{ m/s} \][/tex]
5. Conclusion:
- The speed of the car just before the brakes were applied is [tex]\( 25 \text{ m/s} \)[/tex].
Thus, the speed of the car just before the brakes were applied was [tex]\( 25 \text{ m/s} \)[/tex].
1. Understanding the Problem:
- We are given the total mass of the car: [tex]\( m = 1600 \)[/tex] kg.
- The heat produced when the car is brought to rest: [tex]\( Q = 500 \)[/tex] kJ.
- The car is brought to rest, which implies that its initial kinetic energy was fully converted into heat energy.
2. Conversion of Units:
- Convert the heat produced from kilojoules (kJ) to joules (J). Since [tex]\( 1 \)[/tex] kJ [tex]\( = 1000 \)[/tex] J, we have:
[tex]\[ Q = 500 \text{ kJ} = 500 \times 1000 \text{ J} = 500{,}000 \text{ J} \][/tex]
3. Using the Work-Energy Principle:
- The kinetic energy ([tex]\( KE \)[/tex]) of the car before the brakes were applied is converted into heat energy.
- The formula for kinetic energy is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
- We know that the kinetic energy is equal to the heat energy produced:
[tex]\[ \frac{1}{2} m v^2 = Q \][/tex]
4. Solving for the Initial Speed [tex]\( v \)[/tex]:
- Plug in the known values (mass [tex]\( m \)[/tex] and heat energy [tex]\( Q \)[/tex]):
[tex]\[ \frac{1}{2} \times 1600 \times v^2 = 500{,}000 \][/tex]
- Simplify and solve for [tex]\( v^2 \)[/tex]:
[tex]\[ 800 \times v^2 = 500{,}000 \][/tex]
[tex]\[ v^2 = \frac{500{,}000}{800} \][/tex]
[tex]\[ v^2 = 625 \][/tex]
- Taking the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{625} \][/tex]
[tex]\[ v = 25 \text{ m/s} \][/tex]
5. Conclusion:
- The speed of the car just before the brakes were applied is [tex]\( 25 \text{ m/s} \)[/tex].
Thus, the speed of the car just before the brakes were applied was [tex]\( 25 \text{ m/s} \)[/tex].
