Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the other pulls

Question

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the other pulls in the same direction with a force of 220 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate’s coefficient of kinetic friction on the floor

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Thu Thủy 4 years 2021-08-26T11:21:06+00:00 1 Answers 9 views 0

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  1. Thunguyet
    0
    2021-08-26T11:22:20+00:00
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    Answer:

    \mu_k=0.18

    Explanation:

    First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

    x: T+F-f_k=0\\\\y:N-mg=0

    Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

    Since f_k=\mu_k N and N=mg, we can rewrite the first equation as:

    T+F-\mu_k mg=0

    Now, we solve for \mu_k and calculate it:

    \mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18

    This means that the crate’s coefficient of kinetic friction on the floor is 0.18.

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