Engineering Dynamics - Car Striking a Barrier

[ConnorM]

Homework Statement

A car, assumed to be rigid and having a mass of 800 kg, strikes a barrel-barrier installation without the driver applying the brakes. From experiments the magnitude of the force of resistance F_r, created by deforming the barrels successive, is shown as a function of the vehicle penetration, s. If the car strikes the barrier traveling at a velocity v_c = 60 km/h, determine approximately the distance s to which the car penetrates the barrier.

http://imgur.com/LYc87dW - here is a picture of my question. This has the graph of how the barrels resistance force increases with s.

Homework Equations

v_c = 60 km/h = 16.67 m/s
m = 800 kg

T (Kinetic energy) = 1/2 * mv^2

U_1->2 = ∫F ds

∑U_1->2 = T_2 - T_1

The Attempt at a Solution

Starting off I assumed that once the car had reached the distance s it will have zero kinetic energy. So my equation became

∑U_1->2 = - T_1,

∫F ds = - 1/2 *mv^2, would my bounds of integration be 0 -> s ?

I don't really know what to do, am I using the right principles here?

 

[da_nang]

You got it somewhat figured out. Keep in mind the signs when simplifying this:
$$E_{\text{Final}} - E_{\text{Initial}} = \int \vec{F}(\vec{s}) \cdot \text{d}\vec{s}$$

 

[ConnorM]

OK so then I would have,

∫F ds 0 -> s = - 1/2 *mv2

∫F ds 0 -> s = 111,155.56 J

So from the graph, 40 kN x 0.75m = 30 J

Then I need 111,155.56 J - 30,000 J = 81,155.56 J,

so 60 kN x s = 81,155.56 J

s = 81,155.56 J / 60 kN

s = 1.35 m, then 1.35m + 0.75m = 2.10m

 

[Nick_613]

I did something similar, but I don't get how you end up with 111,155.56 J. -1/2 (800)(60)^2 = 1,440,000. I wanted to go about it the same way as you but as I added up the separate distances, they went past the graph and still didn't get close to the amount of energy needed. Can anyone please clarify this?

... 5 seconds later I realize we use m/s not km/h in SI, after spending two hours trying to figure it out... sorry thanks anyways

 

[canon23]

OK so then I would have,

∫F ds 0 -> s = - 1/2 *mv2

∫F ds 0 -> s = 111,155.56 J

So from the graph, 40 kN x 0.75m = 30 J

Then I need 111,155.56 J - 30,000 J = 81,155.56 J,

so 60 kN x s = 81,155.56 J

s = 81,155.56 J / 60 kN

s = 1.35 m, then 1.35m + 0.75m = 2.10m

I think I'm taking the same dynamics course as you, I finished my CAP assignment yesterday so I might be able to help.

Remember Fr is opposite to the displacement therefore negative such that:

T1 + ∑U1→2 = T2
1/2mv^2 - Area = 0

Also you are measuring area so your units will be according to the graph "kN m" not "J" such that:

Area = 111155.56 N ft = 111.16 kN ft I also got "s = 2.10 m" as my final answer, so I'm assuming it's right.