Your light truck weighs 3,500 kg . You drive up a gentle slope that is 0.5 km long .At the top of the slope yo?

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are 200 m higher in elevation than when you started. Assuming no friction, What force has your engine exerted to drive up this slope? A.7000N B.12000N C.14000N D.20000N

4 Answers

  • The Answer is C. 14000N

    NOTE: you should mention that gravity = 10 MS^-2 and not 9.81 in this question

    You need to resolve forces along the slope, and this requires using the angle of the slope as well as the weight of the truck…essentially you get 2 equations

    M = Mass of Truck, g = Gravity, A= angle of slope, R = reaction of Ground to truck, F = Force of engine

    1) F = Mg SIN(A)

    2) R = Mg COS(A)

    we can ignore equation 2 as it is resolving forces perpendicular to the slope

    SIN(A) = Oppostie side/Hypotenues = 200m/500m = 0.4

    so F = 3500 x 10 x 0.4 = 14000N

    therefore…

    The force exerted by engine = 14000N

    Note: If there was friction the equations would become

    1) F = Mg SIN(A) + UR (U = coefficient of Friction)

    2) R = Mg COS(A)

  • W = F * S

    W = maS

    work done is against gravity .

    W = mgh

    take g = 10

    W =3500*10*200

    =7000000

    W = FS

    F = 7000000/500 =14000N

    Source(s): brain
  • 12000N

  • 14000 obviously.!

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