A woman pushes an oak chest across an oak floor at a constant speed of 0.450 m/s. The chest has a mass of 40.0 kg, and the coefficient of kinetic is 0.620. What is the force exerted by the woman?

A. 243 N
B. 11.1 N
C. 3.81 N
D. 29.0 N

(If there's work please show the best you can)

Respuesta :

lucic

Answer:

A. 243 N

Explanation:

Friction is the force that opposes the relative motion between systems that are in contact.

This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.

First find the normal force which is the force that would point directly upwards to support weight of the block.

Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.

Given m=40 kg and g=9.80 m/s²

N force=40×9.80 =392N

Then find the force of friction which is given by the formula;

F=μN where μ is friction coefficient for the  oak chest  and N is the normal force on the chest

Given μ=0.620  and N force = 392 N then it will be;

F=0.620× 392 =243.04 N

Answer : 243 N