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Ap Physics, Work?

A cart loaded with bricks has a total mass

of 22.9 kg and is pulled at constant speed by a rope. The rope is inclined at 26.2◦above the horizontal and the cart moves 9.9 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.3.The acceleration of gravity is 9.8 m/s2. How much work is done on the cart by the


Answer in units of kJ.

2 Answers

  • Fred
    Lv 7
    4 years ago
    Favorite Answer

    Start by drawing a force diagram!

    The forces on the cart are:

    • friction, horizontally, opposite the direction of motion (-x)

    • gravity, downward (-y)

    • tension in the rope, at angle θ above horizontal (+x, +y)

    • the normal force, upward (+y)

    Since the motion is horizontal in a gravitational field, no work is done against (or by) gravity.

    So the only work done is done against friction.

    And because the motion is at constant velocity, no work goes into kinetic energy.

    And it will be the force of friction (which you'll have to find), times the distance moved (which you're given).

    The force of friction is µ times the normal force:

    F⃗[fric] = µ ֿN

    The only force opposing friction comes from the tension, T, in the rope, and since the horizontal acceleration=0, these opposing forces balance; so the x-component of forces gives:

    Tcosθ = µ N

    The y-component of forces gives:

    N + Tsinθ = Mg

    Now what you're looking for is the work, which is just the force of friction, F[fric], times distance traveled, D.

    D is given, so you need to find F[fric] = µ N.

    But µ is given, so you need to find N.

    And that is done by solving those two equations for T and N; or just for N, if possible:

    N + T sinθ = Mg _ _ _ • cosθ _ _ N cosθ + T sinθ cosθ = Mg cosθ

    T cosθ = µ N _ _ _ _ _ • sinθ _ _ T sinθ cosθ = µ N sinθ

    Subtract, collect N-terms on LHS:

    N cosθ + µ N sinθ = Mg cosθ

    N = Mg cosθ/(cosθ + µ sinθ)


    Work = µ N D = µ Mg D cosθ/(cosθ + µ sinθ)

    = µ Mg D/(1 + µ tanθ)

    = Mg D/(1/µ + tanθ) _ _ _ [this gets each variable to appear just once, making it a bit simpler to key in]

    All those are known, so that's all you need to do!

    [This answer agrees with Steve's.]

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  • Steve
    Lv 7
    4 years ago

    F = µ*m*g/(cosΘ + µsinΘ) = 65.38 N

    Fx = F*cosΘ = 58.66 N

    W = Fx*D = 580.8 J

    Source(s): My 'cheat sheet' of physics formulae
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