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.
* Physics/Math - bobpursley, Saturday, February 10, 2007 at 7:12pm
Break the force F in to vertical and horizontal components.
Friction= (mg + force vertical)mu
net force= ma
horizontal F - force friction = ma
* Physics/Math - COFFEE, Saturday, February 10, 2007 at 9:30pm
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??? PLEASE help .
I have no idea what you did.
Break the foroce F into horisontal components.
Forcevertical F*sintheta
Friction force= mu* (mg + F sinTheta)
Force horizonal= F * cosTheta
Force horizontal net=f*cosTheta-forcefriction
Now set that equal to mass*acceleration. Solve for acceleration
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