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9 August, 13:33

A 5.20-kg package slides 1.41 m down a long ramp that is inclined at 13.0 below the horizontal. The coefficient of kinetic friction between the package and the ramp is mk = 0.313. Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.22 m>s at the top of the ramp, what is its speed after it has slid 1.41 m down the ramp?

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  1. 9 August, 13:34
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    a) Wf = - 21.936 J

    b) Wg = 16.18 J

    c) Wn = 0 J

    d) W = - 5.756 J

    e) v2 = 1.65 m/s

    Explanation:

    Given:

    - The mass of the package m = 5.20 kg

    - The inclination of ramp θ = 13°

    - The coefficient of kinetic friction mk = 0.313

    - The package slides down a distance of s = 1.41 m

    - The initial velocity of the package v1 = 2.22 m/s

    Find:

    (a) the work done on the package by friction;

    (b) the work done on the package by gravity;

    (c) the work done on the package by the normal force;

    (d) the total work done on the package.

    (e) If the package has a speed of 2.22 m>s at the top of the ramp, what is its speed after it has slid 1.41 m down the ramp?

    Solution:

    - Apply Equilibrium conditions perpendicular to the surface:

    N - m*g*cos (θ) = 0

    N = m*g*cos (θ)

    Where, N is the normal contact force exerted by the surface on the package.

    - The frictional force Ff is given by:

    Ff = mk*N

    Ff = - mk*m*g*cos (θ)

    - The work done Wf by frictional force is given by:

    Wf = Ff*s

    Wf = - s*mk*m*g*cos (θ)

    Plug in values:

    Wf = - 1.41*0.313*5.2*9.81*cos (13)

    Wf = - 21.936 J

    - The work done by the contact force Wn is given by:

    Wn = s_y*N

    Since, the equilibrium conditions are maintained perpendicular to surface and no distance is traveled; hence, s_y = 0

    Wn = 0 J

    - The total work done by gravity Wg is given by moving the vertical distance h from start to finish:

    Wg = m*g*h

    Wg = m*g*s*sin (θ)

    Wg = 5.2*9.81*1.41*sin (13)

    Wg = 16.18 J

    - So, the total work done W is given by the sum:

    W = Wf + Wn + Wg

    W = - 21.936 + 0 + 16.18

    W = - 5.756 J

    - Construct an energy balance for the two states at top and bottom of the ramp:

    0.5*m*v1^2 + W = 0.5*m*v2^2

    2 * (v1^2 + W/m) = v2^2

    v2 = sqrt ((v1^2 + 2*W/m))

    Plug in values:

    v2 = sqrt ((2.22^2 - 5.756*2/5.2))

    v2 = sqrt (2.71455)

    v2 = 1.65 m/s
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