Homework Help: Force and Magnitude Problem

1. Oct 9, 2004

VinnyCee

Week long Force and Magnitude Problem

I have this problem that I have been working on for about a weel with no progress:

A car weighs 13,000 N and moves at 40 km / h (11.1 m / s) and it stops in 15 m, the force that stops the car is constant. What are the following?:

a) magnitude of the force stopping the car.
b) time required for stop.

I have tried everything I could think of. What equation shold I be using to figure this one?

For b) I have 15 m = (11.1 m/s) (X) s
then x = 1.35 s, right?

For a) I have (1.35 s) / (11.1 m / s) = 8.22 (13,000 N) = 106,089 N force stopping the car, right? (Probably not, but hopefully soomeone can tell me why.)

Last edited: Oct 9, 2004
2. Oct 9, 2004

arildno

a) Determine the average force from work+energy equation
b) Use that expression for the average force in the impulse+momentum equation to determine t

3. Oct 9, 2004

Diane_

Part a) Think energy. The car has a certain amount of kinetic energy when it's moving. In order for the car to stop, that energy has to go somewhere, and it has to get there somehow. In this case, there will be a force that acts through a distance to perform work on the car - an energy balance with a little division should get you the answer.

b) I'm not sure where some of your numbers are coming from. You seem to be using the

s = ut + (1/2)at^2 equation, but I don't see enough terms to justify that.

Remember: Under uniform acceleration, you can get the average velocity by averaging the initial and final velocities, both of which you know. You also know that the average velocity is displacement over time - you know the displacement and you need the time. So:

s/t = (u + v)/2

t = 2s/(u + v)

Be sure you keep proper track of your units, as they're all mixed up in this.

4. Oct 9, 2004

VinnyCee

Thank you Diane,

For b) I know that Avg. Velocity = displacement (15m) divided by time (unkown). But how do I figure the Avg. Velocity?

5. Oct 9, 2004

Diane_

Again, you can get the average velocity (under uniform acceleration only) by averaging the initial and final velocities. Without a unit change, it'll be 20 kph.

Arildno's solution would work too, of course - it might be worth your while to work it both ways, as you should get the same answer.

6. Oct 9, 2004

arildno

b) You don't need that at all!
Look at your equation of momentum instead:
Let F be the average force found in a).
Then:
$$F*t=m(v_{final}-v_{initial})$$
or:
$$t=\frac{m(v_{final}-v_{initial})}{F}$$

7. Oct 9, 2004

arildno

I agree with Diane on this (just so you don't get confused!)

8. Oct 9, 2004

VinnyCee

Thany you ardilano,

But what is the "work/force" equation that you mentioned for a)?

9. Oct 9, 2004

arildno

Change of kinetic energy equals work done.
Work equal averge force times distance.

10. Oct 9, 2004

VinnyCee

I have recalculated and found the avg. velocity to be (11.1 m/s) / (2) = 5.55 m/s, right?

Then I used that in the formula (what is the name of it?) t = (displacement) / (avg. velocity) to get t = (15 m) / (5.55 m/s) and t = 2.7 s, right?

Then I used the formula F = [m*(finalvelocity - initial velocity)] / t
which would be [13000 N * (0-11.1 m/s)] / (2.7 s) and F = 53,444 N, right?

I think that seems kind of high. where did I go wrong?

11. Oct 9, 2004

arildno

EDIT:
YOU ARE MIXING UP MASS AND WEIGHT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
:grumpy:

12. Oct 9, 2004

VinnyCee

OIC!

Weight = mass * gravity, right?

Then 13000 N = mass * 9.81 m/s^2 and mass = 1325 kg?

And, F = [(1325 kg)(11.1 m / s^2)] / (2.7 s)
F = 14707.5 N / 2.7 s so then F = 5447 N?

13. Oct 9, 2004

arildno

In which direction does your force act, relative to your initial velocity?

14. Oct 9, 2004

VinnyCee

In the opposite direction, right?

so -5447 N?

15. Oct 9, 2004

arildno

Right!..

16. Oct 9, 2004

VinnyCee

Thank you all very much! I don't think I could have figured this out with out you.

17. Oct 9, 2004

arildno

Note:
I haven't bothered to check your numbers, only your method of solving.
This looks right, but you might check the actual numbers just to be sure..

18. Oct 9, 2004

VinnyCee

The units don't match for the last step:

F = [(1325 kg)(-11.1 m / s^2)] / (2.7 s) = (-14707.5 N) / (2.7 s) = -5447 (kg)(m/s)?

19. Oct 9, 2004

arildno

NO, NO, NO!!!!!!!!
:grumpy:
Mass unit: kg
Velocity: m/s
Time:s
(kg*(m/s))/s=(kgm)/s^2=N

20. Oct 9, 2004

VinnyCee

It still does not match (I think):

F = [(1325 kg)(-11.1 m / s)] / (2.7 s) = (-14707.5 (kg)(m/s)) / (2.7 s) = -5447 (kg)(m)?

Last edited: Oct 9, 2004
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