Energy: Finding Maximum spring constant

[MitsuShai]

Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1503 N will move with speed 1.57 m/s at the top of a ramp that slopes downward at an angle 28.7 degrees. The ramp will exert a 721 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 7.7 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.
Calculate the maximum force constant of the spring k_max that can be used in order to meet the design criteria.

Homework Equations

Us=(1/2)kx^2

The Attempt at a Solution

I found the forces:
KE- 189J
PE- 5560J
Wf= -5551.7J
Fw- 721.78
Ff= 721 N

And then:
(1/2)kx^2= (Fw-f)L+KE
(1/2)kx^2= 195J
this is where I got stuck
I've been trying to find x for hours but I had no luck.

 

[tiny-tim]

Hi MitsuShai!

(try using the X² icon just above the Reply box )

… I've been trying to find x for hours but I had no luck.

I'm a little confused …

the question doesn't ask for x.

Anyway, call the distance x, and use the first part of the question to find an equation for k and x.

Then use the second part of the question to find an inequality for k and x.

Eliminate x from the second equation by using the first equation, to get an inequality for k only.

 

[MitsuShai]

Hi MitsuShai!

(try using the X² icon just above the Reply box )I'm a little confused …

the question doesn't ask for x.

Anyway, call the distance x, and use the first part of the question to find an equation for k and x.

Then use the second part of the question to find an inequality for k and x.

Eliminate x from the second equation by using the first equation, to get an inequality for k only.

Hello

yeah the question didn't ask for x but there's two unknown variables x and k
and I don't really get what you're saying...I don't think you can just cancel x because there are two different distances: the distance of the ramp and the displacement distance of the spring, right? because the x in the spring force equation is the distance in how far the spring compressed.

 

[tiny-tim]

Hello MitsuShai!

(just got up :zzz: …)

...I don't think you can just cancel x because there are two different distances: the distance of the ramp and the displacement distance of the spring, right? because the x in the spring force equation is the distance in how far the spring compressed.

The distance of the ramp is given in the question.

You can treat the gravitational PE and the spring PE separately … for the first (and for the work done by friction), you need only the distance of the ramp (which is given), for the second you need only x.

Try what I said …

Anyway, call the distance x, and use the first part of the question to find an equation for k and x.

Then use the second part of the question to find an inequality for k and x.

Eliminate x from the second equation by using the first equation, to get an inequality for k only.

 

[MitsuShai]

Hello MitsuShai!

(just got up :zzz: …)

The distance of the ramp is given in the question.

You can treat the gravitational PE and the spring PE separately … for the first (and for the work done by friction), you need only the distance of the ramp (which is given), for the second you need only x.

Try what I said …

Hi Tim,

There's suppose to be a spring force too, I forgot that...so the spring force should be equal to the force of the crate, so it won't rebound, could the spring force be 1503N?
then
(1/2)kx^2=PE+KE+Wf
why would you need the distance of the ramp?

 

[tiny-tim]

Hi MitsuShai!

(1/2)kx^2=PE+KE+Wf
why would you need the distance of the ramp?

If Wf is the work done by friction, then of course you need the distance.

 

[MitsuShai]

Hi MitsuShai!


If Wf is the work done by friction, then of course you need the distance.

ok, but I solved for Wf with the distance, so do want me to write the equation like this:

(1/2)kx^2=PE+KE+F*d?

I still don't see how x will cancel. I think I'm suppose to calculate x somehow.

 

[MitsuShai]

This problem is so stupid T.T

 

[MitsuShai]

Thank you Tiny Tim for your help. I found how to solve for x now (took me 3 days >:[ ggrrrr).

 

[MitsuShai]

Hello MitsuShai!

(just got up :zzz: …)


The distance of the ramp is given in the question.

You can treat the gravitational PE and the spring PE separately … for the first (and for the work done by friction), you need only the distance of the ramp (which is given), for the second you need only x.

Try what I said …

I actually need your help again, ok I solved the problem before and got the right answer but I don't remember how I did that

After I found all the forces and energies listed above I wrote this down:

.5kx^2=[wsin(theta)-Ff] x distance + KE

and the the equation for the spring constant is
kx= wsintheta + friction force

solve for k on the second, plug into the first, get x, plug that number into the second and you get your k value

I used that equation and I did not get the right answer, which was 5340J.

This is what I did:
kx=wsin(theta)+Ff
k=1442.78/x

then .5kx^2=[wsin(theta)-Ff] x distance + KE
.5(1442.78/x)x^2=[wsin(theta)-Ff] x distance + KE
.5(1442.78)x=[ 5560-721]+189
x= 6.10