# Can someone help me with this problem involving kinetic and potential energy?

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.
asked Feb 3, 2013 in PHYSICS
reshown Feb 3, 2013

Mass of the car,m = 2000 kg

Initial velocity,u= 20 km/h=20*(5/18)=5.55 m/s

Final velocity,v=60 km/h=60*(5/18)=16.67 m/s

Initial elevation,z1=0

Final elevation,z2=120 sin25=50.7 m

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To calculate work done by the engine using work energy theorem

T1+V1=T2+V2+W

W=(T1-T2)+(V1-V2)

=(1/2)m(V1^2-V2^2)+mg(z1-z2)

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

=-124000000 J

=-1240 kJ

Work done on the system is nagative

Work done by the engine is ,1240 kJ

Write 20 km/h in m/s 's.

20 km/h = 20 * (5/18) = 5.56 m/s = Vo.

Similarly, 60 km/h = 60 * (5/18) = 16.67 m/s = Vf.

Distance travelled by a car (d) = 120 m.

Use kinetic equation, i.e, Vf2 = Vo2 + 2ad, to find the acceleration.

Substitute the values Vf = 16.67, Vo = 5.56, and d = 120.

16.67 2 = 5.562 + 2 * a * 120.

2 * a * 120 =  16.67 2 - 5.562

2 * a * 120 =  277.78 - 30.9136

2 * a * 120 =  246.8664

a = 246.8664/240

⇒ a = 1.029 m/s2

Weight of the car (w) = 2000 kg.

Force for the acceleration (F₁) = w * a = 2000 * 1.029 = 2057.22 N.

Force component of the weight along the road (F₂)= w * g * sin(θ) = 2000 * 9.81* Sin(25) = 8291.77 N.

Total force along the road (F) = F₁ + F₂ =  2057.22 N + 8291.77 N = 10349.89 N.

Work done by the engine = Force * Displacement = 10349.89 N * 120 m =1241987.735 Nm.

Work done by the engine is 1241987.735 Nm.