Quick Physics question. Help?
A mass attached to an elastic string is dropped from a height. Given all the values (forces, momentum, drop height, spring constant of string), how may I work out how far the elastic string stretches when the ball is at the bottom of the string? Assume the rope does not reach its limit of elasticity, assume no drag, assume string has no mass.
all I require is a general formula of some sort.
March 2, 2012
13 Comments • Newest first
Ok thanks guys. I think I got it.
http://i233.photobucket.com/albums/ee295/one23abc/SDC10844.jpg <<< thats my working
obviously I just need to finish it off by using the quadratic formula to find an expression for x.
also I think I need to recheck some +/- signs.
[quote=LampShadow]@one23abc: x is the total displacement (stretching) the string experiences for any given time, as long as the string doesn't reach the limits of its elasticity.
But yeah, Hooke's Law for potential energy (which is the integral of the force over distance) is: PE = .5*k*x^2
*edit: Do you know what the spring constant and mass of the object are?[/quote]
yes, they are parameters.
@one23abc: x is the total displacement (stretching) the string experiences for any given time, as long as the string doesn't reach the limits of its elasticity.
But yeah, Hooke's Law for potential energy (which is the integral of the force over distance) is: PE = .5*k*x^2
*edit: Do you know what the spring constant and mass of the object are?
But I thought the x in the formula e = 1/2kx^2 is the extension? Or is it just the middle point of the extended string?
@lampshadow, forces don't really help in finding the extension since the force is changing. That just opens up a whole new can of worms because I might have to differenciate or something (and 'time' is unknown)
You need to use Hooke's Law: F = -kx
Where F is the force acting on the spring, k is the spring constant and x is the displacement the spring experiences.
Assuming it is not a damped oscillation.
From the middle point, the spring will oscillate and the distance between from top to middle, and middle to bottom will be equal. We can put the distance as x. Measuring from the top down, the distance is 2x.
[quote=DeadJesters]I have a physics major from MIT so ignore all these other community college graduates. The answer is (1/2(x+y)m^2h)/(k)^2[/quote]
uh.. explain?
but the maximum length need not be exactly 2x. Wouldn't max length be better notated as (x+y), where y is the extension?
He's setting the spring so its 0 at the top, x in the equilibrium position, and 2x as the maximum length.
[quote=SoloZX]let 2x be the max length the spring will stretch
at the bottom all the gravitational potential energy is elastical
therefore
1/2 k x ^ 2 = mg2x
x =4mg/k I believe
edit: gotta multiply that by 2 caz you wanna know 2x :O[/quote]
wait why 2x?
Google.
It has had all the answers to my physics homework c:
F=kx
F=mg
Thinking... [b] I'm not sure what you would want[/b]. Assuming it takes the full force of gravity, x would be the displacement from where you dropped it.
methinks velocity=0