How to find the force of friction?

Tags:
1. Jun 3, 2015

Matthew M

1. The problem statement, all variables and given/known data
What is the force of friction on the elastic powered car as it rolls accross a smooth surface?
GIVENS
Displacement= 16.25m, Change in time= 22.5s, G=9.8m/s, Mass= 1kg
UNKNOWNS
V1=?, A=? Ff=?

2. Relevant equations
F=MA, A=V/T, Change displacement=((V1+V2)/2)T)

3. The attempt at a solution
First, I used the givens using the change in displacement equation to find V1 which I got 1.4m/s , then i subbed it into A=(V2-V1)/t, I got -31.5m/s ( im not sure why i got a negative? maybe deceleration?) and then i subbed it into F=MA to get -31.5 as my force applied...not sure why this is negative either and not sure where to go from here...

2. Jun 3, 2015

PeroK

I guess the car is decelerating under the force of friction and is no longer being powered by elastic? And the figures given are for the motion from an initial velocity to rest?

You ought to see that $-31.5 m/s$ should be $m/s^2$ and that $31.5$ is way too high a number. You need to check your calculation.

But, yes, the acceleration should be negative and the car is decelerating.

3. Jun 3, 2015

jbriggs444

How did you get V1 again? What change in displacement equation are you talking about with what inputs?
How did you get V2?

4. Jun 3, 2015

Matthew M

Hey, i checked the acceleration, i understand that is a high number, but all the numbers are correct... and yes i know it should be m/s^2 thats just a typo my bad... should I have some sort of conversion for meters --> km or is having the units in m fine? and yes the car is coming to a stop from the initial velocity

5. Jun 3, 2015

Matthew M

V1 was calculated through the use of kinematics equations, the displacement and time are taken from a test trial. the car went 16.25 M in 22.5 seconds and V2 is zero since the car came to a stop

6. Jun 3, 2015

PeroK

So, it started at $1.4m/s$, decelerated at $-31.5 m/s^2$ for $22.5s$ and ended up at rest?

7. Jun 3, 2015

Matthew M

thats what it sounds like yes, it doesnt make much sense to me either...

8. Jun 3, 2015

jbriggs444

If your calculations produce nonsense results then either your calculations are wrong, the assumptions that went into them are wrong or the inputs are wrong.

9. Jun 3, 2015

PeroK

$1.4 \times 22.5 = 31.5$

Are you sure that's the right calculation?

10. Jun 3, 2015

Matthew M

Heres all my calculations
F=MA

F=(1)(-31.5)

F=-31.5N

A=ΔV/T

A=(V2-V1)/T

A=(0-1.4)(22.5)

A=-31.5m/s2 (F)

Δd=((V1+V2)/2)Δt

16 ¼ =((V1+0)/2)22.5

(16 ¼/22.5)(2)=V1

1.4 m/s (f) = V1

11. Jun 3, 2015

Matthew M

AHHH thank you i was supposed to divide! not multiply HAHA thanks for all your help, however from there how do i solve for force of kinetic friction?

12. Jun 3, 2015

PeroK

You already have the relevant equation in your first post. If you know the mass and the acceleration ..

And, force is a vector, so it can be negative too.

13. Jun 3, 2015

Matthew M

so just solving for F=MA gives me my force of friction since its in the opposite direction?

14. Jun 4, 2015

haruspex

Odd that it says 'smooth' surface, since that usually means no friction!
I don't know why they bother telling you it's elastically powered. To solve it, you need to be able to determine the initial speed (whether or not you do so), to do that you have to assume constant acceleration, and for that to be true the elastic must be having no affect.
I advise against finding the initial speed as a number and plugging that into the start of the next calculation. That will accumulate errors. Much better to keep everything algebraic until the end.