Friday

December 19, 2014

December 19, 2014

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

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 ...

Re: Physics/Math - A 3.0 kg block, initially in motion, is pushed along a ...

Physics - A 3.4 kg block is pushed along a horizontal floor by a force of ...

physics - A 3.24 kg block located on a horizontal frictionless floor is pulled ...

physics - I am trying to study for my exam i have tomorrow and this type of ...

Physics - A block of mass 3.42 kg is drawn at constant speed a distance of 1.32 ...