# M3 - Elasticity In Springs

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#1
Hi i am stuck on the two questions attached, any help would much be appricated.

Cheers

Streety
0
15 years ago
#2
1st one
- In equilibrium, extension e = 0.2 m
T = k/l.e = mg
-> k = mg.l/e = mg(0.8/0.2) = 4mg

- When the ball is at P below the equilibrium position x(m)
T = 4mg(x+e)/l
F = mg - T (positive axis is downwards)
mx'' = mg - 4mgx/l - 4mge/l
mx'' = -4mgx/l
x'' = -4gx/l
Let w = rt(4g/l)
At t = 0, v = 0, x = 2 - 1= 1m
So a = 1m, and the equation of x is
x = 1.cos(wt)

- Now calculate the velocity of the ball when the string becomes slack
x = -e = -0.2m
So speed of the ball is v = w. rt(a^2 -x^2) = w. rt(1 - 0.04) = rt(0.96)w

-From this position, you find the speed when it hits the ceiling by using the formula
v^2 = u^2 + 2as
where u = rt(0.96)w, a = -g, s = l = 0.8m 0
15 years ago
#3
1. I think BCHL85 did it correctly, but alternatively you could go the conservation of energy way. It'll save you some time I think.

2.
Draw an isosceles triangle with base 0.18m and height 0.4m. Divide this triangle into two congruent right-angled triangles (base=0.09m). Use pythagoras to find the hyp of both triangles = 0.41m.

Consider the equilibrium position. There's a force F=30N holding the stone down against the tension of both strings. Resolve to get:
2TsinA = 30
sinA = 0.4/0.41 => T=15.375N
T=λx/l => 15.375=λ(0.41-0.15)/0.15 => λ=8.87

Now, after the stone is released consider two positions: the equlibrium position and the position at which the string becomes slack.

@ position 2
EPE=0 (since string is slack)
KE=0.5mv²=0.5(0.05)v²=0.025v²

@position 1
KE=0
EPE=λx²/2l=8.87(0.26)²/(2*0.15)=~2J multiply by 2 = 4J

Conservation of energy gives:
4 = 0.025v²
v = 12.6 m/s

Fixed. 0
#4
The answer for part 3 should be 5.6 m/s and for part 9 12.6m/s 0
15 years ago
#5
Sorry!!! I don't know why I considered PE... The system is horizontal, and not vertical! (I forgot that half-way through.) My second mistake was that I forgot to multiply the EPE by 2 (since we have 2 strings). I edited. 0
15 years ago
#6
1.
When the string is in equlibrium:
T = mg
λx/l = mg
λ = mgl/x = 0.8mg/0.2 = 4mg

At the ground:
KE, PE = 0
EPE = λx²/2l = 4mg(2-0.8)²/2(0.8) = 3.6J

At the point when the string is slack:
EPE = 0
PE = mgh = (2-0.8)mg = 1.2mg (ground is reference line)
KE = 0.5mv²

Conservation of energy gives:
0.5mv² + 1.2mg = 3.6mg
v² = 47.04 m/s

After the string becomes slack, the particle is projected freely under gravity. So:
v² = u² + 2as, with u²=47.04, g=-9.8, s=0.8
v² = 47.04 - 15.68 = 31.36
v = 5.6 m/s
0
#7
DVS U R A STAR!!!

Cheers

Streety
0
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