Question

"A 2000-kg car accelerates from 20 to 60 km/h on an uphill road. The car travels 120 m and the slope of the road from the horizontal is 25 degrees. Determine the work done by the engine."

The answer is 1240 kJ using the First Law of Thermodynamics, but please explain sign conventions of Work and whether or not the engine is excluded from the system... thank you

Answer 1

A

Given

mass is m = 2000 kg , the velocities are v1 = 20 kmph = 20*5/18 m/s = 5.5556 m/s

and v2 = 60 kmph = 60*5/18 m/s = 16.667 m/s

the distace travelled up the hill of inclination is 25 degrees is s = 120 m

as the car is moving up the hill the gravitaional component acts on it

the work done by the engine to reach 120 m of distance and equal to the height h = 120sin theta = 120sin 25= 50.71 m

now the sum of k.e and p.e is equal to the work done by the engine

W = 1/2*m(v1^2 -v2^2)+m*g*h

w = (1/2)(2000)(5.55^2-16.67^2)+2000*9.81*(0-50.7) J

W = -1241820.4 J= 1241.820 KJ

so the work done by the engine is W = 1241.820 kJ