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

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To calculate the initial distance d in a spring compression problem, the gravitational potential energy (mgh) of a mass sliding down an incline is converted into the spring potential energy (1/2kx^2) as it compresses the spring. The mass of 3 kg slides down a 30-degree incline, compressing the spring by 0.2 m. The correct approach involves using the equation mgh = 1/2kx^2, where h is the vertical distance related to d. After calculating, the total distance d is found to be 0.544 m, and the 0.2 m compression must be subtracted to determine the initial separation. The discussion emphasizes the importance of correctly applying energy conservation principles in solving the problem.
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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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