Question

A mover pushes a crate across a wooden floor ata constant speed of 0.75m/s. if the coefficient of static friction for wood on wood is 0.20, how much force does the mover excert on the crate

A) 0.15 N
B) 1.47 N
C) 2.94 N
D) 7.35 N
E) 14.7 N

Answer 1

The force exerted by the mover on the crate is approximately 39.2 N.

So, none of the given options is correct

Step-by-step explanation:

In order to find the force exerted by the mover on the crate, we need to use the coefficient of static friction. The equation for static friction is given by:

fs = μs N

where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

The normal force is equal to the weight of the crate, which is given by:

N = m × g

where m is the mass of the crate and g is the acceleration due to gravity.

Substituting the given values, we have:

N = 20.0 kg × 9.8 m/s2 = 196 N.

Using the coefficient of static friction for wood on wood, which is 0.20, we can calculate the force of static friction:

fs = 0.20 × 196 N = 39.2 N.

Therefore, the mover exerts a force of 39.2 N on the crate.

So, none of the given options is correct