What is the correct solution for the maximum compression of the spring?

[albert12345]

I've tried a few times on this problem but I can't get the right answer...

A box with mass m=1,05 kg slides down a ramp with angle=42.39°. Kinetic friction μ=0,1646 between the ramp and the box. It starts from rest. g=9,806m/s². The box hit's a spring with: k=792 N/m after sliding the distance: s= 2,875m, What is the maximum compression of the spring (x)?

I've tried the following solution based on the law of conservative energy:
(m*g*(s+x)*sin(42,39)) - (μ*m*g*(s+x)*cos(42.39)) = 0.5*k*x²

What am I doing wrong?

Thanks,

 

[Doc Al]

Your solution looks OK to me.

 

[albert12345]

Your solution looks OK to me.

I agree. I came up with the answer: x ≈ 0,203m
The right answer is: 0,0144

 

[Doc Al]

What textbook is this problem from? (Just in case I have it.)

(My answer is close to yours.)

 

[albert12345]

It's from an assignment. I really believe that the answer is wrong, but it has never happened before so...I must have done something wrong.

 

[Doc Al]

I see the problem. Is the spring on the incline? Or on the horizontal section after the incline?

Assume the latter and you'll get the given answer.

 

[tms]

(My answer is close to yours.)

So is mine, if that means anything: 0.211 m.

 

[albert12345]

I see the problem. Is the spring on the incline? Or on the horizontal section after the incline?

Assume the latter and you'll get the given answer.

There is no pic. for the problem, but I get your point. Does it really make that big differernce? And We have never had such problems before, so i think the spring is aligned with the plane

 

[Doc Al]

There is no pic. for the problem, but I get your point. Does it really make that big differernce? And We have never had such problems before, so i think the spring is aligned with the plane

Try it and see for yourself!

 

[albert12345]

Try it and see for yourself!

I really don't think it's right. It's going to be the same since Ug2=the height + the compression of the spring.

 

[Doc Al]

I really don't think it's right. It's going to be the same since Ug2=the height + the compression of the spring.

Assume the spring is horizontally oriented after the mass slides down the incline. You'll get an answer that matches the one you say is correct.

 

[albert12345]

Ok! But I don't understand how I should calculate it :/

 

[Doc Al]

Ok! But I don't understand how I should calculate it :/

Similar to what you did before: you'll use energy conservation. Only this time it's even easier to solve. (Only the incline has friction--assume the flat part after the incline has no friction.)

 

[albert12345]

Similar to what you did before: you'll use energy conservation. Only this time it's even easier to solve. (Only the incline has friction--assume the flat part after the incline has no friction.)

I just tried it! It's weird cause I got the same answer again :/

k1 = (1/2)mv^2 equals U2= (1/2)kx^2

 

[Doc Al]

I just tried it! It's weird cause I got the same answer again :/

k1 = (1/2)mv^2 equals U2= (1/2)kx^2

Try again. Show me an equation similar to what you had in your first post.

 

[albert12345]

I calculated the velocity when the box has slided down the plane:
v= 5,581963937m/s (The speed is just before the box hit's the spring

Then I put in the velocity in the ecuation I wrote earlier, together with the spring constant and then I solved for x

 

[albert12345]

I've tried the same formula I used when I started this topic on another assignment of the same type. It shows that the solution in the assignment was wrong!

IDIOTS! We were right al along :D Thanks for your help guys!

 

[Doc Al]

I've tried the same formula I used when I started this topic on another assignment of the same type. It shows that the solution in the assignment was wrong!

It all depends on whether the spring is on the incline or after the incline. You need a diagram.

Edit: Oops... I think I made an error when I solved the problem the second way (with the spring horizontal). It doesn't give the 'correct' answer. So your suspicion is correct: the answer is wrong no matter how you slice it. Your instructor probably made the same error that I made! That's why I got the 'correct' answer that way. :uhh: