How Do You Calculate the Initial Distance in a Spring Compression Problem?

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Homework Help Overview

The problem involves a mass sliding down an incline and compressing a spring. Participants are tasked with finding the initial distance between the mass and the spring, considering energy transformations between gravitational potential energy and spring potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational potential energy and spring potential energy, with some suggesting to compare initial and final energies. There are questions about the correct interpretation of variables and the setup of equations.

Discussion Status

Some participants have offered guidance on how to approach the problem by comparing energies, while others are clarifying the definitions of variables involved. There is ongoing exploration of the correct method to find the initial separation distance.

Contextual Notes

Participants note the importance of correctly identifying the distance traveled along the incline and the distinction between total distance and the initial separation from the spring. There is also mention of gravitational potential energy changing as the spring compresses.

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Homework Statement



A 3-kg mass starts at rest and slides a distance d down a smooth 30o incline, where it contacts an unstressed spring. It slides an additional 0.2m as it is brought momentarily to rest by compressing the spring (k=400 N/m).

Find the initial separation d between the mass and the spring.


Homework Equations


Work=F*d
Ek=1/2mv^2
Espring=1/2kx^2
v2^2-v1^2=2ad
d=vt
Fg=mg

The Attempt at a Solution



I know that vf=0
and that v1=0
I really need help with v2 but I am not too sure how to go at it.
is this way of finding v correct even though the balls velocity is at an angle:
1/2mv^2=1/2kx^2
v= sqrt16/3 m/s


can anybody please help?
 
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krisrai said:
I really need help with v2 but I am not too sure how to go at it.
is this way of finding v correct even though the balls velocity is at an angle:
1/2mv^2=1/2kx^2
v= sqrt16/3 m/s
I assume v2 is the speed when the ball first contacts the spring. Your calculation is almost right, but you forgot about gravitational PE, which changes as the spring is compressed.

While there's nothing wrong with calculating v2 to solve this problem, you can also solve this problem without finding v2. Just compare initial and final energy.
 
would someone further clarify how to solve this problem?

if you are to compare initial and final energy, should it be:
mgh= 1/2kx^2 ... solve for h?
h= xsin30
 
hi88 said:
if you are to compare initial and final energy, should it be:
mgh= 1/2kx^2 ... solve for h?
Assuming h is measured from the lowest point, yes.
h= xsin30
No. (x is given, by the way.)
 
okay, so here it goes:

mgh=1/2kx^2
(3kg)(9.8)(sin30h)= 1/2(400N/m)(0.2m)
solve for h
h= 2.72 m

correct?


gravitational potential energy being converted to kinetic energy
 
hi88 said:
okay, so here it goes:

mgh=1/2kx^2
Here, h is the vertical distance between the starting point and the lowest point.
(3kg)(9.8)(sin30h)= 1/2(400N/m)(0.2m)
solve for h
h= 2.72 m
I think you mean "d" instead of "h". Two problems:
(1) You forgot to square the 0.2m.
(2) Here, "d" would be the entire distance the mass travels along the incline, not just the initial separation between the mass and the spring (which is what you need to find).
gravitational potential energy being converted to kinetic energy
More like gravitational PE converted to spring PE.
 
correct, d would be the entire distance the mass travels along the incline.. so then, how would you go about finding the initial separation between the mass and the spring?

(3kg)(9.8)(sing30d)=0.5(400n/m)(0.2)^2
d= 0.544 m

would you subtract 0.2 m?
 
hi88 said:
would you subtract 0.2 m?
Yes.
 

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