A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor

Question

A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.

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Minh Khuê 5 years 2021-09-05T04:26:50+00:00 1 Answers 9 views 0

Answers ( )

  1. Edana
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    2021-09-05T04:28:23+00:00
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    Answer:

    Explanation:

    From the information given;

    mass of the crate m = 41 kg

    constant horizontal force = 135 N

    where;

    s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m

    coefficient of kinetic friction u_k = 0.28

    a)

    To start with the work done by the applied force (W_f)

    W_F = F\times (s_1 +s_2) \times cos(0) \ J

    W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J

    Work done by friction:

    W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J

    Work done  by gravity:

    W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j

    Work done by normal force;

    W_n = N \times (s_1 + s_2) \times cos (90) \ J

    W_n = 0 \ J

    b)

    total work by all forces:

    W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12

    W = 2100.5  J

    c) By applying the work-energy theorem;

    total work done = ΔK.E

    W = \dfrac{1}{2}\times m \times (v^2 - u^2)

    2100.5 = 0.5 \times 41 \times v^2

    v^2 = \dfrac{2100.5}{ 0.5 \times 41 }

    v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}

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