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12 April, 00:11

A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity?0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is? k=0.460 and the coefficient of static friction is? s=0.817. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

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  1. 12 April, 01:15
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    d = 11.79 m

    Explanation:

    Known data

    m=4.25 kg : mass of the block

    θ = 37.5° : angle θ of the ramp with respect to the horizontal direction

    μk = 0.460 : coefficient of kinetic friction

    g = 9.8 m/s² : acceleration due to gravity

    Newton's second law:

    ∑F = m*a Formula (1)

    ∑F : algebraic sum of the forces in Newton (N)

    m : mass s (kg)

    a : acceleration (m/s²)

    We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

    Forces acting on the block

    W: Weight of the block : In vertical direction

    N : Normal force : perpendicular to the ramp

    f : Friction force: parallel to the ramp

    Calculated of the W

    W = m*g

    W = 4.25 kg * 9.8 m/s² = 41.65 N

    x-y weight components

    Wx = Wsin θ = 41.65*sin 37.5° = 25.35 N

    Wy = Wcos θ = 41.65*cos 37.5° = 33.04 N

    Calculated of the N

    We apply the formula (1)

    ∑Fy = m*ay ay = 0

    N - Wy = 0

    N = Wy

    N = 33.04 N

    Calculated of the f

    f = μk * N = 0.460*33.04

    f = 15.2 N

    We apply the formula (1) to calculated acceleration of the block:

    ∑Fx = m*ax, ax = a : acceleration of the block

    -Wx-f = m*a

    -25.35-15.2 = (4.25) * a

    -40.55 = (4.25) * a

    a = (-40.55) / (4.25)

    a = - 9.54 m/s²

    Kinematics of the block

    Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block:

    vf²=v₀²+2*a*d Formula (2)

    Where:

    d:displacement (m)

    v₀: initial speed (m/s)

    vf: final speed (m/s)

    dа ta:

    v₀ = 15 m/s

    vf = 0

    a = - 9.54 m/s²

    We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d)

    vf²=v₀²+2*a*d

    0 = (15) ²+2 * (-9.54) * d

    2 * (9.54) * d = (15) ²

    (19.08) * d = 225

    d = 225 / (19.08)

    d = 11.79 m
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