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2 years ago

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A 51.1-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.622 and 0.368,

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:

Charra [1.4K]2 years ago

4 0

Answer:

(a) The force must be greater than 311.4851 N to just start the crate moving

(b) The 184.287 N force needed to to slide the crate at constant speed

Explanation:

We have given that mass of the crate m = 51.1 kg

Coefficient of static friction $\mu _s=0.622$ and coefficient of kinetic friction $\mu _k=0.368$

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

(A) Static frictional force is given by $F_S=\mu _Smg=0.622\times 9.8\times 51.1=311.48516N$

So the force must be greater than 311.4851 N to just start the crate moving

(b) For sliding the crate across the dock at a constant speed force must be equal to kinetic friction force

Kinetic friction force is given by $F_k=\mu _kmg=0.368\times 9.8\times 51.1=184.287N$

So the 184.287 N force needed to to slide the crate at constant speed

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