Need quick help with Univ. Physics I problem

[Sanonuke22]

So I took my physics final this morning, only to find out that I have a missing lab somehow, that happens to be required to pass. I took a look at the problems, and as embarrassed as I am to say it, I do not have any idea of how to work the problem. Problem is, I have to have it in by tomorrow morning. Any help would be appreciated!Lab Problems:

1. Given a 30 degree ramp with a frictional coefficient = 0.10 and a spring with k = 1000N/m
that pushes a 100 gram projectile up the ramp, how long of a distance must the spring be
compressed to launch the 100 gram projectile at a 100m/s velocity? Assume that the ramp is
just long enough for the spring to fully release its compressed energy at the end of the ramp.

2. What percentage of the energy was lost due to friction? In your opinion was it worth taking
it into account? Why or why not?

3. What if we had the same spring/ramp system but increased the frictional coefficient to 1.0,
and dropped the launch velocity to 20m/s? What would be the increase in frictional losses as
a percentage of total energy in the system?

What I've tried so far:

(1/2)kx^2 = sin(30)mgx + "mew"mgx + (1/2)mv^2

then used the quadratic formula to get x = 1.0006 or -0.9994

 

[ehild]

What I've tried so far:

(1/2)kx^2 = sin(30)mgx + "mew"mgx + (1/2)mv^2

then used the quadratic formula to get x = 1.0006 or -0.9994

That term "mew"mgx for the work against friction is not correct. What is the force of friction on a slope? Your equation would be correct otherwise.

You find Greek letters to the right under "Quick Symbols", just click on them. "mew" is μ.



ehild

 

[Sanonuke22]

Hey, thanks for the help. (1/2)kx^2 = sin(30)mgx + μmgcos(30) + (1/2)mv^2

How does this look?

 

[ehild]

Hey, thanks for the help. (1/2)kx^2 = sin(30)mgx + μmgcos(30) + (1/2)mv^2

How does this look?

Much better - perfect!

ehild

 

[Sanonuke22]

Here's what I got if anyone wants to do it as well so I can compare:

a) 1.000545 m

b) 0.017% - No ; E lost due to friction is negligible in this case

c) x = 2.047 m ; 0.04%

 

[ehild]

Well, there is a mistake in the equation, (I did not notice previously): you left out the x in the friction term. (1/2)kx^2 = sin(30)mgx + μmgcos(30)x + (1/2)mv^2.

ehild

 

[Curious3141]

That term "mew"mgx for the work against friction is not correct. What is the force of friction on a slope? Your equation would be correct otherwise.

You find Greek letters to the right under "Quick Symbols", just click on them. "mew" is μ.



ehild

Sorry, ehild, just couldn't resist.

 

[ehild]

It is soooo cute!

But it is not a Hungarian cat. The cats say "mew" but μ does not sound like "mew" here

ehild

 

[Curious3141]

It is soooo cute!

But it is not a Hungarian cat. The cats say "mew" but μ does not sound "mew" here

ehild

Don't know about Hungary cats. But in Singapore, the only cats that go "mew" are Hungry Cats.

OK, back to the serious part of the thread. Sorry, OP.