Velocity related to force of friction

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Homework Help Overview

The problem involves a 5kg object sliding down a frictionless inclined plane and then moving along a horizontal surface with kinetic friction. The objective is to determine the time it takes for the object to stop after leaving the inclined plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy to find the kinetic energy as the object leaves the inclined plane. There are attempts to relate kinetic energy to the work done by friction and to derive the time taken to stop.

Discussion Status

The discussion includes various interpretations of the equations involved, with some participants questioning the validity of certain approaches. There is acknowledgment of confusion regarding the relationship between velocity and work done by friction, and some participants are clarifying their statements and correcting their equations.

Contextual Notes

Participants are navigating through the implications of kinetic energy and friction, with some expressing uncertainty about the setup and the equations used. There is a focus on ensuring the correct interpretation of energy conservation principles in the context of the problem.

UrbanXrisis
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A 5kg object slides 100m down a frictionless inclined plane dropping 45m. It then slides along a horrizontal surface with a coefficiet of kinetic friction of .75 intil it stops. How long does it take to stop after it leaves the inclined plane?

I'm not sure how to get the time. Here's what I found:

mgh=.5mv^2
10m/s/s*45m=.5 v^2
v=30 m/s

So the object is going 30m/s when it hits the friction floor. I'm not sure what to do next.
 
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Hi there,

This is a conservation of Energy question.

The most important thing to do is to set up the question right.

We now that the Ek = Ethermal (Ethermal is the energy lost to the fristional force)

Ek= squrt(2gh) because the Ek will be resulting from the drop on the inclined plane and the 45m drop.

Ethermal= mgd0.75

squrt(2gh)=mgd0.75
Solve for distance then if you the velocity just use V=d/t to find the time!
 
To determine the kinetic energy of the object as it leaves the inclined plane, use conservation of energy.

Now you have the kinetic energy given by [itex]E_k = \frac{1}{2}mu^2[/itex]. Get the value of [itex]u[/itex] and then plug it into [itex]v = u + at[/itex] where [itex]v = 0[/itex] and [itex]a = - \frac{F_{fr}}{m} = - \mu g[/itex] and solve for [itex]t[/itex].

EDIT : I misread the question, they asked for "how long" (time) not distance. :blushing:
 
Last edited:
I'm sorry,but your last equation doesn't make any sense.You equate velocity to work done by friction force...:wink:

Daniel.
 
dextercioby said:
I'm sorry,but your last equation doesn't make any sense.You equate velocity to work done by friction force...:wink:

Daniel.

You're talking about my post or derek's ?
 
You hadn't written your reply,by the time i got into writing mine,so that's why i didn't quote the mistake.There it is:
derekmohammed said:
squrt(2gh)=mgd0.75

Daniel.
 
Opps. :smile:

I just finished a two hour lecture on aristotle :zzz: when I wrote this post.

I see what I did I am :rolleyes: sooooo sorry. What I really meant was that EK = Ep and that you could find Velocity through that Equation... Sorry. the final Equation should be:

mgh = mgd0.75

Where h is the total distance the box falls!

SORRY
 
Last edited:

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