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

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.

Answer 1

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

 

Answer 2

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.

 Source : https://answers.yahoo.com/question/index?qid=20130202124423AA1Pxmc