Posted by **Stephanie** on Saturday, October 15, 2011 at 12:55pm.

When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?

- physics -
**Henry**, Sunday, October 16, 2011 at 6:16pm
Wb = mg = 1.7kg * 9.8N/kg = 16.66N.

Fb = (16.66N,0deg) = Force of book..

Fp = 16.66sin(0) = 0 = Force parallel to table.

Fv = 16.66cos(0) = 16.66N. = Force perpendicular to table.

1. Fap - Ff = 0, Fap = Force applied.

2.60 - Ff = 0, Ff = Force of friction.

Ff = 2.60,

u*Fv = 2.60,

16.66u = 2.60,

u = 0.156. = Static coefficient of friction.

2. 1.5 - Ff = 0,

Ff = 1.5,

u*Fv = 1.50,

16.66u = 1.5,

u = 0.09 = Kinetic coefficient of friction.

## Answer this Question

## Related Questions

- physics - When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to ...
- physics - When you push a 1.55 kg book resting on a tabletop it takes 2.10 N to ...
- physics - When you push a 1.55 kg block resting on a tabletop it takes 1.85 N to...
- physics - When you push a 2.05 kg book resting on a tabletop it takes 2.45 N to ...
- physics - When you push a 1.95 kg book resting on a tabletop it takes 2.40 N to ...
- Physics - When you push a 1.82-kg book resting on a tabletop, it takes 2.30 N to...
- physics - I keep geting this question wrong and cant figure out what i am doing ...
- phy - When you push a 1.82-{\rm kg} book resting on a tabletop, it takes 2.21 N ...
- Physics - When you push a 1.80-kg book resting on a tabletop, it takes 2.05 N to...
- Physics - boy pushes a 3.3-kg book against a vertical wall with a horizontal ...