Question

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 and 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?

Answer 1

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

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is $F_{s} =\mu_{s} mg$.

but once crate starts moving the force of friction is reduced  $F_{v} =\mu_{k} mg$.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie $F_{v} =\mu_{k} mg$.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

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