Finding the coefficient friction of static and kinetic force...

Question

a force of 50.0N is required to start a 6.0kg box to move across a horizontal concrete floor.

A. what is the coefficient of the static friction?

B. if the 50.0N force continues, the box accelerated to 0.80m/s^2. what is the coefficient of kinetic friction?

show the solution please so that i may be able to further understand the problem.

Answer 1

coefficient of static friction =limiting force of static friction/normal force

=50/normal reaction;where normal reaction =mass x gravity

gravity =9.8or 10m/s2

=50/6 x 10

=50/60=0.833333333

dunno bt kinetic force...........but coefficent of kinetic friction=forceof kinetic friction/normal reaction..............

jus check if itz rite.................

Answer 2

Vibha is correct for the static coefficient:

Force to overcome friction = 50 N = mus * m * g, so

mus = 50 / 6 / 9.8 = 0.85

For the kinetic coefficient you need to use F=ma, where

F=50 N (say to the right)

Ff = friction force resisting motion (acts to the left) = muk * m * g



It's the net force (F - Ff) that is causing the given accelleration (0.8), NOT the 50 N force), so

F - Ff = m * a

F - muk * m * g = m * a; you know all except muk, so solve for muk:

muk = (F - m * a) / (m * g)

muk = (50 - 6 * 0.8) / (6 * 9.8)

muk = 0.77

Which is reasonable since usually mus > muk.

Good Luck!