Homework Help

Find the coefficient of kinetic friction. A dock worker loading crates on a ship finds...

user profile pic

shawn123 | Student, Undergraduate | eNotes Newbie

Posted October 27, 2009 at 9:46 PM via web

dislike 1 like

Find the coefficient of kinetic friction.

A dock worker loading crates on a ship finds that a 15kg crate, initially at rest on a horizontal surface, requires a 64N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 57N is required to keep it moving with a constant speed.

The acceleration of gravity is 9.8 m/s^2.

2 Answers | Add Yours

user profile pic

krishna-agrawala | College Teacher | Valedictorian

Posted October 27, 2009 at 10:36 PM (Answer #1)

dislike 1 like

Given:

Mass of crate = m = 15 kg

Force required to set the crate in motion = f1 = 64 N

Force required to keep the crate in moving at constant speed = f2 = 57 N

Acceleration of gravity = g = 9.8 m/s^2

We know:

Coefficient of kinetic friction =

(Force required to keep the crate in moving at constant speed)/(Perpendicular force exerted on the crate)

And:

Perpendicular force exerted on the crate = m*g

Substituting given value of m and g we get:

Perpendicular force exerted on the crate =15*9.8 = 147 N

Substituting this value of perpendicular force and Force required to keep the crate in moving at constant speed in equation for coefficient of kinetic friction we get:

Coefficient of kinetic friction = 57/147 = 0.3878

user profile pic

neela | High School Teacher | Valedictorian

Posted October 27, 2009 at 10:38 PM (Answer #2)

dislike 1 like

Kinetic friction is the maximum force required to maintain the uniform  motion or  maintain the constant velocity. Under such condition , the net force due to the kinetic frictional force and the exerted force is zero.

The kinetic frictional force = Mk * mg, where Mk is the coeefficient of kinetic friction between the crate and the horizontal surface.m is the mass of the crate and g is the acceleration due to gravity.

Applied force =57 N,

Therefore, the net force = 0.

Mk*(15*g  N)+57N=0 Therefore, Mk= -57N/(15*9.8)N = -0.3878  or   the coefficient of friction is 0.3878.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes