Calculate the work done by friction

AI Thread Summary
The discussion revolves around calculating the work done by friction on a cart with a mass of 1.12 kg, initially moving at 0.57 m/s and slowing to 0.427 m/s over 2 seconds. The Work-Energy Theorem is applied, leading to the calculation of work as W = 1/2 m (Vf^2 - Vi^2), resulting in a value of -0.0798 kg m/s, indicating negative work done by friction. Participants clarify that this negative work reflects friction's role in slowing the cart. Additional questions arise about using distance and delta K to find friction force, confirming the use of the Work-Energy Theorem for such calculations.
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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