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11 September, 19:40

A box of mass 15.1 kg with an initial velocity of 2.33 m/s slides down a plane, inclined at 23◦ with respect to the horizontal. The coefficient of kinetic friction is 0.56. The box stops after sliding a distance x. How far does the box slide? The acceleration due to gravity is 9.8 m/s 2. The positive x-direction is down the plane. Answer in units of m

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  1. 11 September, 21:23
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    x=2.21m

    Explanation:

    A box of mass 15.1 kg with an initial velocity of 2.33 m/s slides down a plane, inclined at 23◦ with respect to the horizontal. The coefficient of kinetic friction is 0.56. The box stops after sliding a distance x. How far does the box slide? The acceleration due to gravity is 9.8 m/s 2. The positive x-direction is down the plane. Answer in units of m

    Energy lost by box = work done by friction.

    If the box stops after distance x, it has lost potential energy of mgx. sinθ and Kinetic energy of (1/2) mVo^2.

    F=μN

    coefficient of kinetic friction is the ratio of frictional force to the normal reaction

    The normal reaction from the plane is mgcosθ,

    friction force = μmg. cosθ, and

    since work done is the product of force and distance covered

    the work done by friction is μmgx. cosθ.

    So mgx. sinθ + (1/2) mVo^2 = μmgx. cosθ

    (1/2) mVo^2 = mgx (μ. cosθ - sinθ)

    x = Vo^2/2g (μ. cosθ - sinθ)

    slotting the parameters into the equation above, we have

    m=15.1kg

    velocity=2.33m/s

    angle θ=23deg

    The coefficient of kinetic friction is 0.56.

    the acceleration due to gravity g = 9.8 m/s^2

    x=2.33^2 / (2*9.81 (0.56. cosθ23 - sin23))

    x=2.21m
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