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20 April, 03:48

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.83 m above the bottom of the chute with an initial speed of 1.91 m/s. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.269. How far from the bottom of the chute does the toy zebra come to rest

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  1. 20 April, 05:18
    0
    Given that,

    height h = 1.83 m

    Initial speed Vi = 1.91m/s

    Coefficient of kinetic friction

    uk = 0.269

    According to the law of conservation of energy, states that sum of the potential energy and kinetic energy at top of the chute is equal to the kinetic energy at bottom.

    ∆P. E = ∆K. E

    mgh = ½m (Vf²-Vi²)

    m cancel out

    gh = ½Vf² - ½Vi²

    ½Vf² = gh + ½Vi²

    Vf² = 2gh + Vi²

    Vf² = 2*9.81 * 1.83 + 1.91²

    Vf² = 39.5527

    Vf = √39.5527

    Vf = 6.29 m/s

    The frictional force exerted on the toy from the bottom of the chute is

    Fr = - uk•N

    N = W = mg

    Fr = ma

    Then, ma = - uk•m•g

    Divide through by m

    a = - uk • g

    a = - 0.269 * 9.81

    a = - 2.64 m/s²

    According to equation of motion to find the distance travel at the bottom

    v² = u² + 2as

    Note: the final velocity is zero, the body comes to rest. And the initial velocity is the velocity when the zebra reached the ground u = Vf = 6.29m/s

    Then, v² = u² + 2as

    0² = 6.29² + 2 (-2.64) s

    -6.29² = - 5.278s

    - cancels out

    Then, s = 6.29²/5.27

    s = 7.5m

    The zebra moves 7.5m from the chute
  2. 20 April, 05:28
    0
    The zebra comes to rest at 4.58 m from the bottom of the chute

    Explanation:

    We are given;

    Height; h = 1.83 m

    Initial speed; Vi = 1.91 m/s

    Coefficient of kinetic friction; = 0.269

    Now, from the law of conservation of energy,

    Initial Kinetic Energy + Initial Potential Energy = Work done by friction

    Now,

    Kinetic Energy is given by the formula, KE = (1/2) mv²

    Potential Energy is given by the formula; PE = mgh

    Also, work done by friction is given as W = µmgd

    Thus, we now have;

    ½mv² + mgh = µmgd

    m will cancel out and we have;

    ½v² + gh = µgd

    Plugging in the relevant values to get;

    ½ (1.91) ² + (9.8 x 1.83) = (0.44 x 9.8) d

    1.82405 + 17.934 = 4.312d

    d = 19.75805/4.312

    d = 4.58 m
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