Posted by **COFFEE** on Saturday, February 10, 2007 at 5:56pm.

A 3.0 kg block, initially in motion, is pushed along a horizontal floor by a force F of magnitude 18 N at an angle = 45° with the horizontal. The coefficient of kinetic friction between the block and floor is 0.25. (Assume the positive direction is to the right.) Calculate the magnitude of the frictional force on the block from the floor. Calculate the magnitude of the block's acceleration.

Break the force F in to vertical and horizontal components.

Friction= (mg + force vertical)mu

net force= ma

horizontal F - force friction = ma

Ok, so this is what I tried...

T = uk*m*g / ((cos theta)+uk(sin theta))

T = ((.25)(3)(9.8)) / ((cos -45)+(.25)(sin -45))

T = 13.9 N

then,

FNet = m*g - T*sin(theta)

FNet = (3)(9.8) - (13.9)*(sin -45)

FNet = 39.2

then,

fk = uk*FNet

fk = .25*39.2

fk = 9.8

Well that was the wrong answer but I don't see anything wrong with my approach. Any suggestions???

## Answer This Question

## Related Questions

- physics - A 3.5 kg block, initially in motion, is pushed along a horizontal ...
- physics - A 3.5 kg block, initially in motion, is pushed along a horizontal ...
- - A 3.5 kg block, initially in motion, is pushed along a horizontal floor by a...
- physics - A 3.0 kg block is pushed along a horizontal floor by a force F of ...
- Physic - A 3.5 kg block is pushed along a horizontal floor by a force of ...
- Physics - A 3.4 kg block is pushed along a horizontal floor by a force of ...
- Re: Physics/Math - A 3.0 kg block, initially in motion, is pushed along a ...
- AP Physics C - A 4.10 kg block is pushed along a floor by a constant applied ...
- physics - A 3.24 kg block located on a horizontal frictionless floor is pulled ...
- physics - A 4.30 kg block is pushed along a floor by a constant applied force ...

More Related Questions