Spring/distance ball problem

krisrai · May 22, 2008

1. The problem statement, all variables and given/known data

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.

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

3. The attempt at a solution

I know that vf=0
and that v1=0
I really need help with v2 but im 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?

Doc Al · May 22, 2008

krisrai said: ↑

I really need help with v2 but im 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.

hi88 · May 28, 2008

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

Doc Al · May 28, 2008

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.)

hi88 · May 28, 2008

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

Doc Al · May 28, 2008

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.

hi88 · May 28, 2008

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?

Doc Al · May 28, 2008

hi88 said: ↑

would you subtract 0.2 m?

Yes.

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Spring/distance ball problem

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Doc Al

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hi88

Doc Al

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hi88

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hi88

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