Physics Help Please!?

Question

When you push a 1.70 kg book resting on a tabletop it takes 2.40 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?

Static?

Kinetic?

If you can just tell me to go about how to solve the problem that would be fine. (Though answers are always welcome to.) I am drawing a complete blank.

Answer 1

the static force will be equal to 2.4 N, such that the normal force (m*g) times the static coefficient is equal to 2.4.

Likewise, since there is no acceleration
1.5=m*g*kinetic coefficient.

j

Answer 2

The equation we need is:

Frictional force = coefficient of friction * normal force

The normal force is the force that is perpendicular to the surface the object is moving on (the book, in this case). Since there is no vertical motion, the normal force cancels out the weight of the object, hence

normal force = weight of object. (be careful here with mass and weight again)

the equation will always work whether you are looking for the static or kinetic coefficients of friction

The static coefficient is always larger than the kinetic because it requires a larger force to get something moving that to keep it moving.

Hope this helps. :)

Answer 3

static friction is the friction on the bottom surface of that book and the object before you can push it in motion.

(notation - v means subscript)

fvs < or = to (uvs) (N)

where uvs is the coefficient of static friction and N is the normal force (which in this case is mg). Note that it is < or equal to, since friction would dissipate the force you apply (no matter how small it is) and the book will not move unless you reach and apply a force greater than fvs's.

once you apply a force greater than fvs's, the book will move and will have a lesser coefficient of friction, fvk. (notice how easier it is to keep a heavy cabinet moving than starting it from rest).

fvk= (uvk) (N)

where uvk is the coefficient of kinetic friction.

since the object is in equilibrium (constant speed), F net =0,
thus fvk+Fapplied on book=0

fvk=-1.50N

just solve by plugging in values. hope this helps.

Answer 4

since it takes 2.4 N to start the book sliding,then F=sR where F=force needed to overcome friction=2.4N,s=coefficient of static friction & R=normal reaction force=1.7N. Therefore s=F/R=2.4/1.7=1.412. Again,when it is sliding with constant speed,there is kinetic friction. Here F=kR where k=coeffn of kinetic friction.F=1.5N &R=1.7N.So k=F/R=1.5/1.7=0.882.