Does Friction Affect the Acceleration of a Sliding Block?

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Homework Help Overview

The discussion revolves around a physics problem involving the effects of friction on a sliding block. The problem includes calculating the minimum force required to initiate movement and understanding the block's motion once it is sliding, considering both static and kinetic friction coefficients.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of forces involved, questioning the relationship between applied force and friction. There is an attempt to clarify the conditions under which the block accelerates or moves at constant velocity.

Discussion Status

Some participants have made progress on part of the problem, specifically in calculating the force needed to overcome static friction. Guidance has been offered regarding the application of Newton's second law, but uncertainty remains about the subsequent motion of the block.

Contextual Notes

Participants are working under the constraints of the problem statement, including specific values for mass and friction coefficients, and are navigating the implications of these values on the block's motion.

inkandgold
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Homework Statement


The coefficient of static friction between a block and a horizontal floor is 0.44, while the coefficient of kinetic friction is 0.10. The mass of the block is 5.4 kg and it is initially at rest.

(a) What is the minimum horizontal applied force required to make the block start to slide?
(b) Once the block is sliding, if you keep pushing on it with the same minimum starting force as in part (a), does the block move with constant velocity or does it accelerate?
(c) If it moves with constant velocity, what is its velocity (in m/s)? If it accelerates, what is its acceleration (in m/s2)?


Homework Equations


F = ma
fs = usN


The Attempt at a Solution



a) F=ma
F=(5.4kg)(9.8m/s/s) = 52.92, which isn't right

b) I believe it will accelerate.

c) It's not 9.8 m/s/s. I will probably have to find it via F=ma?

Thanks so much for your help!
 
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Well, I figured out part a! a = 5.4 kg * 9.8m/s/s *.44 = 23.3. Still unsure about the rest though.
 
inkandgold said:
Well, I figured out part a! a = 5.4 kg * 9.8m/s/s *.44 = 23.3. Still unsure about the rest though.
You can answer the rest by using Newton's 2nd Law, Fnet = ma, where Fnet is the algebraic sum of the applied force and the friction force. Watch plus/minus signs.
 
Thanks so much, got it now!
 

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