A crate with a mass of m = 130 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate and the deck is As-0.86. The coefficient of kinetic friction is μk-0.41. Randomized Variables m 130 kg μ,-0.86 Pk=0.41 25% Part (a) Write an expression for the minimum force, F,n, that must be applied to get the block moving from rest. Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining:5 (5% per attempt) detailed view 0 Submit Part (b) What is the magnitude of the force F,n in newtons? Part (c) Write an expression for the force F, that must be applied to keep the block moving at a constant velocity Part (d) what is the magnitude of the force F., in newtons?

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Xema8 Xema8 · Answer 1 · 2020-02-19T04:47:09+01:00

Answer:

Explanation:

a,b )

Minimum force required to move the crate

= maximum static friction

= μs x mg

= 0.86 x 130 x 9.8

= 1095.64 N

c,d ) Force required to keep the crate in uniform motion or net force is zero.

F = maximum kinetic friction

= μk x mg

= .41 x 130 x 9.8

= 522.34 N

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tallinn tallinn · Answer 2 · 2022-02-16T16:43:23+01:00

(a, b )

The Minimum force required to move the crate

Then = maximum static friction

After that = μs x mg

Then = 0.86 x 130 x 9.8

Now = 1095.64 N

(c, d ) Force required to keep the crate in uniform motion or net force is zero.

F = maximum kinetic friction

After that = μk x mg

= .41 x 130 x 9.8

Thus, = 522.34 N

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