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General Physics problem using kinematics and dynamics, application of newton's law - friction kinetic friction coefficient

Posted on 2008-10-02
Medium Priority
Last Modified: 2008-10-02
A 4.5-kg block slides down an inclined plane that makes an angle of 28 degress with the horizontal. Starting from rest, the block slides a distance of 2.4m in 5.2 seconds. Find the coefficient of kinetic friction between the block and plane.

This is a problem from chapter 5 of Paul A. Tipler's Physics for Scientists and Engineers Chapter 5. I have the answer (odd problem, 121) and have spent at least a half hour trying to arrive there.

This problem is multistep and requires the use of (I assume) X = X_initial + V_initial * time + .5 * a * time^2
, net force = mass * acceleration , and   Force_friction = coefficient * Force_normal. Sorry I do not know how to use subscripts and superscripts or greek letters with this site.

I am thinking:
Step 1: Find the normal force (force perpendicular to the plane the block is on),     m*gravity*sin(28)?
Step 2: Find acceleration (down the plane),  see equation above, I came up with .1775m/s/s
Step 3: Find magnitude of force of friction, the normal force perpendicular to the plane - the friction force = m * a
Step 4: Lastly, use the firctional force, and the perpendicular force to find the ceofficient of kinetic friction from the equation       F_fr = coefficient * perpendicular force

I keep coming up with with friction and perpendicular forces that are both ~39N, which would lead to coefficient of 1, which is by far not the answer. If it helps anyone the correct answer is .51

But how do we get there, and is there somethign wrong with my approach or am I missing a bad calculation?
Question by:justinvoels
LVL 18

Accepted Solution

deighton earned 2000 total points
ID: 22626134
I get that the perpendicular force is 4.5 * g * cos(28)=38.977

the accelerating force on the block is 4.5*g*sin(28)=20.752

the acceleration is .1775, so force left doing the acceleration is m a = 4.5*.1775=.7988

so frictional force = 20.752-.7988=19.953



Author Comment

ID: 22626286
I made an officially lame mistake, I was hitting sine for both instead of cosine for perpendicular force. I figured it was such a mistake, but it makes me feel better I had the method correct. Thanks!

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