Proper design of automobile braking systems must account for heat buildup under heaving braking. Calculate the thermal energy dissipated from brakes in a 1360-kg car that descends a 13.5° hill. The car begins braking when its speed is 88 km/h and slows down to a speed of 36 km/h in a distance of 0.5 km measured along the road.
To calculate the thermal energy dissipated from the brakes, we need to determine the work done by the braking force and the change in potential energy of the car as it descends the hill.
1. Determine the change in potential energy:
The change in potential energy is given by the formula:
ΔPE = mgh
where m is the mass of the car (1360 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical height of the hill.
To find the vertical height of the hill, we can use trigonometry:
h = d * sin(θ)
where d is the distance along the road (0.5 km) and θ is the angle of the hill (13.5°).
Converting distance to meters:
d = 0.5 km = 0.5 * 1000 m = 500 m
Converting angle to radians:
θ = 13.5° * (π/180) = 0.2356 radians
Plugging in the values:
h = 500 m * sin(0.2356) ≈ 118.29 m
ΔPE = 1360 kg * 9.8 m/s² * 118.29 m = 1,712,966.24 J
2. Determine the work done by the braking force:
The work done is given by the formula:
W = F * d
where F is the braking force and d is the distance along the road.
Converting speeds to m/s:
v₁ = 88 km/h = 88/3.6 m/s ≈ 24.44 m/s
v₂ = 36 km/h = 36/3.6 m/s ≈ 10 m/s
Plugging in the values:
d = 0.5 km = 500 m
The average acceleration during braking can be calculated using the formula:
a = (v₂² - v₁²) / (2 * d)
Plugging in the values:
a = (10 m/s)² - (24.44 m/s)² / (2 * 500 m) ≈ -2.45 m/s²
The braking force can be calculated using Newton's second law:
F = m * a
Plugging in the values:
F = 1360 kg * -2.45 m/s² ≈ -3342 N (negative sign indicates opposite direction of motion)
The work done by the braking force is then:
W = -3342 N * 500 m = -1,671,000 J
3. Calculate the thermal energy dissipated:
The thermal energy dissipated is equal to the sum of the change in potential energy and the work done by the braking force:
E = ΔPE + W
Plugging in the values:
E = 1,712,966.24 J + (-1,671,000 J) = 41,966.24 J
Therefore, the thermal energy dissipated from the brakes in this scenario is approximately 41,966.24 Joules.