# Calculate the work done by friction

• cc2hende

## 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

Welcome to PH.

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?

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.