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: \(\( m = 1600 \)\) kg.
- The heat produced when the car is brought to rest: \(\( Q = 500 \)\) 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 \(\( 1 \)\) kJ \(\( = 1000 \)\) J, we have:
\(\[
Q = 500 \text{ kJ} = 500 \times 1000 \text{ J} = 500{,}000 \text{ J}
\]\)
3. Using the Work-Energy Principle:
- The kinetic energy (\(\( KE \)\)) of the car before the brakes were applied is converted into heat energy.
- The formula for kinetic energy is:
\(\[
KE = \frac{1}{2} m v^2
\]\)
- We know that the kinetic energy is equal to the heat energy produced:
\(\[
\frac{1}{2} m v^2 = Q
\]\)
4. Solving for the Initial Speed \(\( v \)\):
- Plug in the known values (mass \(\( m \)\) and heat energy \(\( Q \)\)):
\(\[
\frac{1}{2} \times 1600 \times v^2 = 500{,}000
\]\)
- Simplify and solve for \(\( v^2 \)\):
\(\[
800 \times v^2 = 500{,}000
\]\)
\(\[
v^2 = \frac{500{,}000}{800}
\]\)
\(\[
v^2 = 625
\]\)
- Taking the square root of both sides to solve for \(\( v \)\):
\(\[
v = \sqrt{625}
\]\)
\(\[
v = 25 \text{ m/s}
\]\)
5. Conclusion:
- The speed of the car just before the brakes were applied is \(\( 25 \text{ m/s} \)\).
Thus, the speed of the car just before the brakes were applied was \(\( 25 \text{ m/s} \)\).
