(a) A truck of mass 1870 kg travelling at 80 km/h braked to 55 km/h then continued at that speed. How much work was done by the brakes during that period?
I just wanted to check my answer... 243334 J ??
To calculate the work done by the brakes, we need to find the change in kinetic energy of the truck when it goes from 80 km/h to 55 km/h.
The kinetic energy (KE) of an object can be calculated using the formula KE = (1/2) * mass * velocity^2.
First, let's convert the speeds from km/h to m/s, as our formula requires velocity in meters per second (m/s).
80 km/h = 80 * (1000 m / 1 km) / (3600 s / 1 h) = 22.22 m/s
55 km/h = 55 * (1000 m / 1 km) / (3600 s / 1 h) = 15.28 m/s
Next, calculate the initial kinetic energy (KE_initial) and final kinetic energy (KE_final):
KE_initial = (1/2) * mass * velocity_initial^2
= (1/2) * 1870 kg * (22.22 m/s)^2
KE_final = (1/2) * mass * velocity_final^2
= (1/2) * 1870 kg * (15.28 m/s)^2
Now, find the difference in kinetic energy (ΔKE) by subtracting KE_final from KE_initial:
ΔKE = KE_final - KE_initial
Finally, the work done by the brakes is equal to the change in kinetic energy, so:
Work = ΔKE
Calculate the result using the given values:
Work = ΔKE
= KE_final - KE_initial
= (1/2) * 1870 kg * (15.28 m/s)^2 - (1/2) * 1870 kg * (22.22 m/s)^2
Now, let's calculate the value to confirm your answer:
Work ≈ 243,348 J
Therefore, your answer of 243,334 J is very close to the correct value. Keep in mind that rounding errors or variations in the formula used might account for this slight discrepancy.