How to use the work-energy theorem to find mu?

[jshot079]

Homework Statement

please check this link out to make it simpler to understand...

http://www.scribd.com/doc/62694749/MC-Collision1D2

there is a friction block and a spring scale... it gives two measurements (the vertical force and the horizontal force)
F_vertical=88.2 N and
F_horizontal=75.8 N
now I'm suppose to find the coefficient of friction...
and also, I'm suppose to find out if the driver exceeded the 70 km/h speed limit
please help me out!

Homework Equations

m₁v_i1+m₂v_i2 = (m₁+m₂)v_f

KE=1/2(m)(v²)

The Attempt at a Solution

for the mu:
at first, i tried to use the "elementary" method of summation of forces along the axes but since we're talking about collisions, i think i should use the work-energy theorem... but now i don't know how to put mu into the equation: KE=1/2(m)(v²)
...

for the speed limit:
well, it's an inelastic collision obviously... one car is at rest but they didn't give out the v_i1 of the yellow car...

please help! i ran out of ideas on how to solve this!

 

[Redbelly98]

Welcome to Physics Forums. I'm going to copy-and-paste the problem statement, just because that makes it easier to discuss here.

You are just starting a summer job for PDRM collision investigators. You assist theinvestigators in order to determine, as accurately as possible, the circumstancessurrounding collisions (initial velocity of the vehicles involved, direction vehicles wereheading, etc.) in order to take action against drivers at fault under the Criminal Code(dangerous driving causing death or other charges). The job is demanding and your help ishighly appreciated. The weather is great, it’s July, and youarrive at the station as your boss tells you,“Hey, Rookie, hurry over to Kangar Street. There has been an accident involving twocars. There are people injured.” Listeningonly to your courage, you rush (althoughcarefully) to the scene of the accident. Your job is to take photos of the vehicles as theyare positioned, and take a certain number of measurements, including the length of theskid marks and layout of the debris. Youimmediately draw the following sketch of the accident and include a few common measurements.

Left: Vehicle 1 starts braking - - - - - Center: Impact - - - - - Right: Final position of vehicles
- - - - - 30 m skid marks for Vehicle 1 braking - - - - - 12.25 m skid marks after impact

Your findings are as follows:
•Numerous debris (broken glass, plastic, etc.) were found at a distance of 12.25 mfrom the cars;
•The two cars stuck together and there are skid marks over 12.25 m;
•There are also skid marks over a distance of 30 m before the debris;
•The posted speed limit on this street is 70 km/h;
•Other data: m₁ = 2.674 kg, m₂= 1.100 kg

Homework Statement

please check this link out to make it simpler to understand...

http://www.scribd.com/doc/62694749/MC-Collision1D2

there is a friction block and a spring scale... it gives two measurements (the vertical force and the horizontal force)
F_vertical=88.2 N and
F_horizontal=75.8 N
now I'm suppose to find the coefficient of friction...

Okay. What is an equation that involves friction force and coefficient of friction?

and also, I'm suppose to find out if the driver exceeded the 70 km/h speed limit
please help me out!

Homework Equations

m₁v_i1+m₂v_i2 = (m₁+m₂)v_f

Good, that equation is the basis for solving problems like this.

KE=1/2(m)(v²)

The Attempt at a Solution

for the mu:
at first, i tried to use the "elementary" method of summation of forces along the axes but since we're talking about collisions, i think i should use the work-energy theorem... but now i don't know how to put mu into the equation: KE=1/2(m)(v²)
...

for the speed limit:
well, it's an inelastic collision obviously... one car is at rest but they didn't give out the v_i1 of the yellow car...

They don't give v_i1 because that is what you are supposed to find out.

For starters, I would:

1. Figure out what μ, the coef. of friction, is. You can use the measurements F_vertical=88.2 N and F_horizontal=75.8 N, along with the usual equation that involves friction force and μ.

2. Figure out how fast the two cars were going after the collision. You can use the 12.25 stopping distance, and what the acceleration must have been given the μ calculated previously.

 

[jshot079]

thanks for confirming and killing any doubts i have.

for the mu, i used the equation for the spring scale and friction block


µ = Fspring /Fnormal
µ =Fspring /(mblock •g )

so i assumed that the horizontal force is from the spring scale and the vertical force is from the friction block...

so i got:

= 75.8/88.2
µ = 0.86

i don't think I'm right... my physics professor said that i should use the work and KE theorem to solve for mu... how am i suppose to do that?

 

[Redbelly98]

You are correct on µ. It sounds like your professor misspoke (or, possibly, was misheard? ). Work and KE theorem can be used to find velocities that are relevant to this problem, but that involves knowing µ already -- which now you do.

m1vi1+m2vi2 = (m1+m2)vf

What are the unknown quantities here? The next step is to work on finding their values, and the KE theorem will be useful.

 

[jshot079]

hahah... tnx! I knew i was right... one way or another.
with that out of the way...

so now i got
m₁=2.674 kg
m₂=1.100 kg
v_i2=0
and with two unknowns...

v_i1=?
v_f =?

and in order to fine them i got these two equations...
v_i1=(m1+m2)vf/m1 and
v_f= m1vi1/(m1+m2)

hmm... would i be wrong if i substituted 70m/s to via? if i am... is there any way to find via and vf at the same time with putting mu and distance into these equations?

 

[Redbelly98]

v_i1=(m1+m2)vf/m1

Yes, so if you can find vf then you'll have v_i1.

hmm... would i be wrong if i substituted 70m/s to via? if i am... is there any way to find via and vf at the same time with putting mu and distance into these equations?

Yes, 70 m/s would be wrong, and for two reasons. (1) It's 70 km/hr, not m/s, but more importantly (2) that is the posted speed limit, and we can't assume the driver's velocity was at the speed limit -- we'd already have our answer then, wouldn't we?

Instead, apply the work-energy theorem to find vf.

Work-energy theorem: W = F_net·d = ΔKE

Apply that to the final 12.25 m distance that the two vehicles travel before coming to a stop.

 

[jshot079]

oh my God, thank you!
i just had that Eureka moment with your help!
XD

 

[jshot079]

ok!
So this is what i did...

well first you need to find F_net so...
its
Ʃf_x
F_a-F_f=+ma

since the velocity is constant... i think that acceleration is zero... so i get

75.8-F_f=0
________________________________
and next...
ƩF_y
F_N-F_w=0
F_N=F_w
F_N=88.2

and if we apply the coefficient of friction to the Ʃf_x... we still get zero so it checks...
_____________________
so our F_net=88.2

using that we get...
W = F_net·d
W = 88.2(12.25)
W = 1080.45
________________________
then applying the energy-work theorem wherein W=ΔKE
ΔKE=1/2mv²
1080.45=1/2(2.674)(v²)
1080/1.337 = v²
808.12=v²
v= 28.43

Right?

Thank you for not giving up on me and for not giving me the answers directly and letting my brain do some work...

so all in all... i won't be pressing charges because he did not exceed the speed limit... it's the red car's fault for stopping anyway... heheh

 

[jshot079]

but before doing that... i should watch the units more carefully next time...

yeah... i will have to charge the yellow car with over speeding... i should really know my units...

so i converted 70 km/hr to m/s

and i get 19.44m/s
and the driver did... get over that limit...

sorry, sometimes I'm a klutz...

 

[Redbelly98]

ok!
So this is what i did...

well first you need to find F_net so...
its
Ʃf_x
F_a-F_f=+ma

since the velocity is constant... i think that acceleration is zero... so i get

75.8-F_f=0

Let me stop you here.

What is this force F_a? Is it a force that acts on the cars? If you haven't already, draw yourself a force diagram that describes the cars skidding 12.25 m after the collision.

After the collision, the cars are going at speed vf, then the come to a stop (v=0) in 12.25 m. So the speed changes, it is not constant.

ƩF_y
F_N-F_w=0
F_N=F_w
F_N=88.2

88.2 N is the weight of the friction block. The cars are different than that, right?

 

[jshot079]

doh! me and my hastiness...
i will work on this!