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

Posted on 2008-10-02
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
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
LVL 18

Accepted Solution

deighton earned 500 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!

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

How to Win a Jar of Candy Corn: A Scientific Approach! I love mathematics. If you love mathematics also, you may enjoy this tip on how to use math to win your own jar of candy corn and to impress your friends. As I said, I love math, but I gu…
This article provides a brief introduction to tissue engineering, the process by which organs can be grown artificially. It covers the problems with organ transplants, the tissue engineering process, and the current successes and problems of the tec…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templa…

756 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question