Derivations of specific values in Physics

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Homework Help Overview

The original poster seeks guidance on deriving specific values in physics, particularly the coefficient of static friction (μs) given a distance of 1.2 km traveled in 17 seconds. The context involves understanding the relationships between distance, time, and acceleration in a physics problem related to motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need for additional information to solve the problem, particularly whether the object comes to rest after the given distance. There are inquiries about the necessary constant acceleration and how it relates to the average velocity calculated from the distance and time.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the relationships between average velocity, acceleration, and the equations of motion. However, there is no explicit consensus on the approach to take or the assumptions that should be made.

Contextual Notes

There is a noted lack of clarity regarding the complete problem statement, which may hinder the ability to derive the necessary values. Participants emphasize the importance of understanding the relationships between the variables involved in the problem.

Ben Nibarger
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Homework Statement


I am wondering if anyone knows of a document that shows derivations of specific values in Physics? For example, I need to find μs and am given the distance traveled and time (1.2 km in 17 sec).

Homework Equations


I know the equation for μs, Ffrictions⋅FN.

The Attempt at a Solution


I can use this to calculate velocity, etc., but I am at a loss of how to proceed and my textbook doesn't have an example. Mastering Physics doesn't provide a "walk me through a similar problem" feature.
 
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You need to know more than that to solve the problem. I suspect the piece of information you left out is that after these 17 seconds (and 1.2 km) the object ends up at rest?
In the future (and perhaps in your next post) you should copy the whole problem statement, preferably word for word.

In physics, there aren't formulas for every kind of problem; it is up to you to put the pieces together.
I'll help you get started: The equation you wrote for the force of friction says that the force is constant, which means that the acceleration is constant.
 
Ben Nibarger said:
I can use this to calculate velocity, etc., but I am at a loss of how to proceed and my textbook doesn't have an example. Mastering Physics doesn't provide a "walk me through a similar problem" feature.
Physics is the art of learning how to solve those problems if you are not given the formula to plug in values (that's the part a computer can do).

There has to be more information to solve the problem. Does the object stop at the end?
 
mfb said:
Physics is the art of learning how to solve those problems if you are not given the formula to plug in values (that's the part a computer can do).

There has to be more information to solve the problem. Does the object stop at the end?

This is the question: Assuming a constant acceleration and no slipping of tires, estimate the coefficient of static friction needed for a drag racer to cover 1.2km in 17s , starting from rest.
 
so, what "constant acceleration" would be needed?
 
Nathanael said:
You need to know more than that to solve the problem. I suspect the piece of information you left out is that after these 17 seconds (and 1.2 km) the object ends up at rest?
In the future (and perhaps in your next post) you should copy the whole problem statement, preferably word for word.

In physics, there aren't formulas for every kind of problem; it is up to you to put the pieces together.
I'll help you get started: The equation you wrote for the force of friction says that the force is constant, which means that the acceleration is constant.

I understand that, I just don't have a strong grasp of the relationships yet. If there was a tool that showed the relationships, or derivations, then it would be easier to understand.
 
lightgrav said:
so, what "constant acceleration" would be needed?

4.15 m/s2
 
Ben Nibarger said:
4.15 m/s2
How do you get this?
 
vavg = Δx/Δt = 1200m/17sec = 70.6 m/s

v = v° + at => v/t = a => 70.59 m/s / 17 s = 4.15 m/s2
 
  • #10
You have the equation "Δv/Δt = v/t = a" ... but Δv ≠ vavg. (You used vavg for the Δv). To do it it your way you have to note that vfinal = 2vavg

This is how I think about it:
vavg = 0.5(vfinal+v0) = 0.5(vfinal) = 0.5(a)(Δt) ... [This only applies when v0=0]
vavgΔt = Δx
Put the first equation into the second equation and you get:
0.5(a)(Δt)2 = Δx ... [Again, it only applies when v0=0]Once you get the correct acceleration, you have to figure out what is causing this acceleration. (What is the net force on the car?) Draw a free-body-diagram.
 
Last edited:

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