A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N that acts p

Question

A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N that acts parallel to the ramp.The coefficient of kinetic friction between the ramp and the incline is = 0.30. If the suitcase travels 4.60 m along the ramp, calculate:A) the work done on the suitcase by force F.B) the work done on the suitcase by the gravitational force.C) the work done on the suitcase by the normal force.D) the work done on the suitcase by the friction force.E) the total work done on the suitcase.F) if the velocity of the suitcase is zero at the bottom of the ramp,what is its velocity after it has travelled 4.60 m along the ramp.

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Adela 3 years 2021-08-24T23:01:18+00:00 1 Answers 21 views 0

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  1. mit
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    2021-08-24T23:02:20+00:00
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    Answer:

    A) 667 J

    B) 381.4 J

    C) 0 J

    D) 245.4 J

    E) 40.2J

    F) 2 m/s

    Explanation:

    Let g = 9.81 m/s2

    A) The work done on the suitcase is the product of the force applied and the distance travelled:

    w = Fs = 145 * 4.6 = 667 J

    B) The work done by gravitational force the dot product between the gravity vector and the distance vector

    W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J

    C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

    D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

    W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J

    E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

    W = w - W_g - W_f = 667 - 381.4 - 245.4 = 40.2 J

    F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

    E_k = W = 40.2 j

    mv^2/2 = 40.2

    20v^2/2 = 40.2

    10v^2 = 40.2

    v^2 = 4.02

    v = \sqrt{4.02} = 2 m/s  

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