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A 2.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.60, and the coefficient of kinetic friction is µk = 0.40. A horizontal force, F, is applied to the block. As F is increased, the block begins moving. Describe how the force of friction varies as F increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of F―before the block starts moving, at the point it starts moving, and after it is moving. Show your work. 

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  • NCS
    Lv 7
    2 weeks ago
    Favorite Answer

    At all times that the block is stationary,

    static friction force fs  = applied force F

    which maintains horizontal equilibrium.

    At the threshold of sliding,

    fs = F = µs*m*g = 0.60 * 2.0 * 10m/s² = 12 N

    using nice round numbers.

    Once it begins moving, the friction force suddenly drops to

    fk = µk*m*g = 0.40 * 2.0kg * 10m/s² = 8.0 N

    using nice round numbers, and so unless F is modified, there will be a net force on the box causing it to accelerate.

    Hope this helps!

  • 2 weeks ago

    A mass of 2kg weighs 2kg×9.8N/kg = 19.6 Newtons 

    Static frictional force is .6×19.6 = 11.76 Newtons 

    Dynamic frictional force is. 4×19.6 = 7.84 Newtons 

    Update if you need more.

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