Posted by **megan** on Tuesday, January 18, 2011 at 4:34am.

A 3.20 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal.

The block accelerates to the right at 7.50 m/s2.

Determine the coefficient of kinetic friction between block and ceiling.

Would appreciate some guidance on how the Normal force works in this case

- physics -
**Henry**, Wednesday, January 19, 2011 at 7:52pm
Fb = 3.2kg * 9.8 = 31.36N = ver(normal)

force of block.

Fh = 85*cos55 = 48.75N.

Fv = 31.36 + 85*sin55 = 101N = ver. force.

Fn = Fh - uFv = ma,

48.75 - 101u = 3.2 * 7.5 = 24,

48.75 -101u = 24,

-101u = 24 - 48.75,

-101u = -24.75,

u = -24.75 / -101 = 0.245 = coefficient

of friction.

NOTE: Fn = Net force.

## Answer this Question

## Related Questions

- Physics - A 4.4 kg block is pushed along the ceiling with a constant applied ...
- physics - A 4.8 kg block is pushed along the ceiling with a constant applied ...
- physics - A 4.10 kg block is pushed along the ceiling with an constant applied ...
- physics - A 3.40 kg block is pushed along the ceiling with an constant applied ...
- physics - A 3.60 kg block is pushed along the ceiling with an constant applied ...
- physics - A 3.40 kg block is pushed along the ceiling with an constant applied ...
- PHYSICS - A 4 kg block is pushed along the ceiling with a constant applied force...
- physics - A 5.55 kg block covered in sandpaper is pushed along the ceiling of a ...
- physics - A 5.55 kg block covered in sandpaper is pushed along the ceiling of a ...
- physics - Two blocks are pushed along a horizontal frictionless surface by a ...

More Related Questions