A 1600.0 kg car with a speed of 75.0 km/h brakes to a stop. How many cal of heat are generated by the brakes as a result?
To calculate the heat generated by the brakes, we need to use the principle of conservation of energy. The kinetic energy of the car is converted into heat energy through the braking process.
First, let's convert the car's mass and speed into SI units:
Mass (m) = 1600.0 kg
Speed (v) = 75.0 km/h = 75.0 km/h × (1000 m/km) / (3600 s/h) = 20.83 m/s
The kinetic energy (KE) of the car is given by the equation:
KE = (1/2)mv^2
Substituting the values:
KE = (1/2) × 1600.0 kg × (20.83 m/s)^2
Now, we need to calculate the change in kinetic energy as the car comes to a stop. Since the car is braking to a stop, its final speed (vf) will be zero.
Change in kinetic energy = KEfinal - KEinitial
= 0 - [(1/2) × 1600.0 kg × (20.83 m/s)^2]
Next, we need to calculate the amount of heat generated. When the brakes are applied, the kinetic energy transforms into heat energy. The conversion factor between kinetic energy and heat energy is 1 calorie (cal) = 4.18 Joules (J).
Let's assume the entire change in kinetic energy is converted into heat energy. Multiply the change in kinetic energy by the conversion factor to get the heat generated:
Heat generated = Change in kinetic energy × Conversion factor
= [(1/2) × 1600.0 kg × (20.83 m/s)^2] × 4.18 J/cal
Calculating this equation will give you the amount of heat generated by the brakes in calories.