# Friction and Energy Conversion

• alison16
In summary, a car traveling at 24 m/s leaves a patch of mud with a 1.7x10^4 net resistive force and has a speed of 23.6 m/s.f

## Homework Statement

A 1100kg car is traveling at 24m/s before it hits a 18m patch of mud with a 1.7 x 10^4 net horizontal resistive force. What is the cars speed as it leaves the mud patch?

## Homework Equations

Change in Uint= fd
KE= 1/2mv^2
W= Fd

## The Attempt at a Solution

1/2mv^2 = f(mud on car)d
1/2(1100)v^2 = 1.7 x 10^4(18)
V= 23.6 m/s

I'm pretty sure this isn't correct, because the resistive force is too great to have the final velocity change only slightly. I feel like I might need to incorporate work or the initial velocity into my equation, but I don't really know how.

## Homework Statement

A 1100kg car is traveling at 24m/s before it hits a 18m patch of mud with a 1.7 x 10^4 net horizontal resistive force. What is the cars speed as it leaves the mud patch?

## Homework Equations

Change in Uint= fd
KE= 1/2mv^2
W= Fd

## The Attempt at a Solution

1/2mv^2 = f(mud on car)d
1/2(1100)v^2 = 1.7 x 10^4(18)
V= 23.6 m/s

I'm pretty sure this isn't correct, because the resistive force is too great to have the final velocity change only slightly. I feel like I might need to incorporate work or the initial velocity into my equation, but I don't really know how.

The car starts out with some initial KE due to its initial velocity. The friction "robs" energy from this initial store of KE. So find out how much KE is left after traversing the mud patch...

Must be the battery in the calculator. Left and right are both of the order of 300000 J so there must be a considerable slowdown.
Means your intuition is good. Both work (from resistive force) and inital velocity (*) are in your equation (assuming your KE means ΔKE), which is just fine. Some typo on the calculator ?

 (*) realize I misinterpreted what you wrote. As gneill (the doctor) says: ΔKE instead of ½mv2 and you're fine.

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Got it! 4.4 m/s is much more resonable. Thank you.