What is the minimum work needed to push a 1225 kg car 340 m up along a 17.5° incline?

Question

anonymous · Answer

(a) Ignore friction.
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(b) Assume the effective coefficient of friction retarding the car is 0.20?
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anonymous · Answer

Turn that incline into a right triangle and then you just have to calculate the vertical height the car actually climbs:

sin 17.5 = opposite side / hypotenuse
0.30 = opposite side / 340m
opposite side = 0.30 * 340m = 102.2m

the opposite side on a right triangle is the y-axis or the vertical height

height = 102.2m

Work = force * distance
work = KE = PE = potential energy of car if it fell from 102.2m down to ground by force of gravity

PE = mgh
PE = 1225 kg * 9.8 m/s^2 * 102.2m

Work = PE = 1,226,911 Joules = 1227 kJ