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3 years ago

11

A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

1 answer:

NISA [10]3 years ago

3 0

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of static friction $\mu _s=0.584$

Coefficient of kinetic friction $\mu _k=0.399$

(a) Static friction force is given by $F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N$

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by $F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N$

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Answer:

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Answer:

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Explanation:

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$V=V_{o}-gt$ ----------------------- (3)

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x = m is the horizontal distance travelled by the golf ball

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