Friction and Energy Conversion

[alison16]

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.

 

[gneill]

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...

 

[BvU]

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 ?

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

 

[alison16]

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

 

[HalfLostAndSearching]

Your attempt at a solution is a good start, but there are a few things that need to be considered in order to get the correct answer.

Firstly, you are correct in thinking that the initial velocity and work done by the resistive force need to be taken into account.

To calculate the final velocity of the car, we need to use the equation:

KE_initial + W_resistive = KE_final

Where KE_initial is the initial kinetic energy of the car, W_resistive is the work done by the resistive force, and KE_final is the final kinetic energy of the car.

We can calculate the initial kinetic energy of the car using the equation KE= 1/2mv^2, which gives us:

KE_initial = 1/2(1100)(24)^2 = 316800 J

Next, we need to calculate the work done by the resistive force. We can use the equation W= Fd, where F is the net resistive force and d is the distance over which the force is applied. In this case, the distance is 18m, so we have:

W_resistive = (1.7 x 10^4)(18) = 306000 J

Now, we can plug these values into our original equation to solve for the final kinetic energy:

316800 + 306000 = KE_final

KE_final = 622800 J

Finally, we can solve for the final velocity by rearranging the equation KE= 1/2mv^2 to solve for v:

v = √(2KE/m)

v = √(2(622800)/1100) = 26.6 m/s

So the final velocity of the car as it leaves the mud patch is 26.6 m/s. This is a decrease from the initial velocity of 24 m/s, which makes sense because the resistive force is slowing the car down.