The coefficient of kinetic friction between a 780.-newton crate and a level warehouse floor is 0.200. calculate the magnitude of the horizontal force required to move the crate across the floor at constant speed.

you must first make a free body diagram of the block to place all the forces that act on the block there. You will have four forces. The normal force, the weight, the horizontal force in x (which is what they ask to calculate) and the friction force. Then you raise the equations and you clear the values. I attach the solution.

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snehashish65 snehashish65 · Answer 2 · 2021-10-12T05:04:12+02:00

Answer:

The magnitude of the horizontal force required to move the crate is 156 N.

Explanation:

Given data:

Weight of crate is, [tex]W = 780 \;\rm N[/tex].

Coefficient of friction is, [tex]\mu =0.200[/tex].

The horizontal force required to move the crate across the floor is equal to the frictional force. Then,

horizontal force = frictional force

[tex]F=f\\F=\mu \times N[/tex]

Here, N is the normal reaction force, which is due to weight of crate (W).

Solving as,

[tex]F=\mu \times W\\F=0.200 \times 780\\F=156 \;\rm N[/tex]

Thus, the magnitude of the horizontal force required to move the crate is 156 N.

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