Posted by **Mackenzie ** on Wednesday, January 4, 2012 at 3:07pm.

A 20 kg table initially at rest on a horizontal floor requires a 136 N horizontal force to set it in motion. Once the table is in motion, a 71 N horizontal force keeps it moving at a constant velocity. What is the coefficient of kinetic friction between the table and floor?

- Physics -
**Henry**, Wednesday, January 4, 2012 at 7:53pm
Wt = mg = 20 kg * 9.8 N/kg = 196 N. =

Weight of table.

Ft = 196 N. @ 0 Deg. = Force of table.

Fp = 196*sin(0) = 0 = Force parallel to floor.

Fv = 196*cos(0) = 196 N. = Force perpendicular to the floor.

Fn = Fap-Fp-Fk = ma = 0, a = 0.

71-0-Fk = 0,

Fk = 71 N. = Force of kinetic friction.

u*Fv = 71,

196u = 71,

u = 0.362 = Kinetic coefficient of friction.

## Answer this Question

## Related Questions

- physics - A 29 kg chair initially at rest on a horizontal floor requires a 375 N...
- physics - A 92 kg clock initially at rest on a horizontal floor requires a 656 N...
- Physics - 1.A 24 kg crate initially at rest on a horizon floor requires a 75 N ...
- Physics - a 96 kg clock initially at rest on a horizontal floor requires a 605 N...
- physics - A 95kg box initially at rest on a horizonal floor requires a 650N ...
- physics - A 25 kg chair initially at rest on a horizontal floor requires 393 N ...
- physics - A 25 kg chair initially at rest on a horizontal floor requires 360 N ...
- Physics - A 25 kg chain initially at rest on a horizontal floor requires a 365 N...
- physics - A 23 kg chair initially at rest on a horizontal ﬂoor requires a...
- physics - A 23 kg chair initially at rest on a horizontal ﬂoor requires a...

More Related Questions