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blsea
3 months ago
15

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

nd the deck is μs = 0.73. The coefficient of kinetic friction is μk = 0.59
Write an expression for the force Fv that must be applied to keep the block moving at a constant velocity.
What is the magnitude of the force Fv in newtons?
Physics
1 answer:
Sav [3.1K]3 months ago
4 0

Answer:F_{v} =\mu_{k} mg

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is F_{s} =\mu_{s} mg.

Once the crate begins to move, the frictional force decreases to F_{v} =\mu_{k} mg.

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to F_{v} =\mu_{k} mg.

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N

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