Calculate the work done by friction

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Homework Help Overview

The problem involves calculating the work done by friction on a cart being pushed along a level track. The cart has a specified mass and initial and final velocities are provided for a time interval.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Work-Energy Theorem and the relationship between work and kinetic energy changes. Questions arise regarding whether the calculated work represents the work done by friction or the cart itself.

Discussion Status

Some participants have provided hints and clarifications regarding the interpretation of the work done by friction. There is an ongoing exploration of different approaches to calculate work based on varying information, such as distance and changes in kinetic energy.

Contextual Notes

Participants are considering different scenarios for calculating work, including cases where distance and changes in kinetic energy are known. The discussion reflects uncertainty about the correct application of formulas and the definitions involved.

cc2hende
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Homework Statement


You have a level track. You push a cart with mass = 1.12[kg].
You measure the initial velocity to be 0.57[m s-1].
2 seconds later, you measure the velocity to be 0.427[m s-1].


Homework Equations


What is the work (reported in mJ) that friction did on the cart?


The Attempt at a Solution



W(f)=F(f) x d
 
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You need to show us your work so we can help you.

Hint: What does the Work-Energy Theorem say?
 
W= deltaK = 1/2m(Vf^2 - Vi^2)

W=1/2 (1.12kg) [((0.427m/s)^2) - ((0.57m/s)^2)]

W= -0.0798 kg m/s

Is this the work done by friction though? or the cart?
 
On what object would the cart do work?

Friction slows the cart, so it does negative work on the cart, so your answer seems reasonable.
 
Okay that makes sense. thank you.

So if i were to calculate work done by friction, but this time I'm given distance and delta K, would I use Wncf=fd where f=(deltaK)/d or work energy theorem?
 
cc2hende said:
Okay that makes sense. thank you.

So if i were to calculate work done by friction, but this time I'm given distance and delta K, would I use Wncf=fd where f=(deltaK)/d or work energy theorem?

That gives you the friction (force) itself.
 

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