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.

Update:

COULD SOMEONE ANSWER!!!!

4 Answers

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  • kasab
    Lv 7
    8 years ago
    Favorite Answer

    20 km/h = 5.56 m/s, and 60 km/h = 16.67 m/s.

    Use, Vf^2 = Vo^2 + 2*a*S

    (16.67 m/s)^2 = ( 5.56 m/s)^2 + 2*a*120 m ------> a = 1.029 m/s^2.

    Force for acceleration = 2000 kg * 1.029 m/s^2 = 2058.1275 N

    Force component of the weight along the road = 2000 kg * 9.81 m/s^2 * Sin(25 deg) = 8291.77 N.

    Total force along the road = 2058.1275 N + 8291.77 N = 10349.8978 N.

    Work done by the engine = Force * Displacement = 10349.8978 N * 120 m =1241987.735 Nm

  • Yes, someone could answer and I would have except you posted that question at least 5 times and then got demanding when you didn't get an immediate answer.

    http://answers.yahoo.com/activity;_ylt=Ah2nVL3Cg3y...

    And you do it with just about every question you have. Knock it off, kiddie.

    We're not your answer b%$ches.

    I wish Kasab had not answered. He is just rewarding your awful behavior.

    Do you know that Kasab's answer may not be correct?

    Source(s): 30 years engineering and know a self-entitled slacker when I hear one.
  • Bomba
    Lv 7
    8 years ago

    There will be two components of work involved - the one associated with linear acceleration of the mass plus the one associated with the lifting of the weight. So does the car weigh 2000 kg or is 2000 kg the car mass ?

  • 4 years ago

    the two could have comparable replace as by using the regulation of capability conservation =>KE(preliminary) = PE(very final) & the m2 will income greater top than m1 as KE(preliminary) is comparable for the two=>v2>v1 =>h2>h1

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