The Student Room Group
Shickles
Let me just reproduce the question here
A jack-in-the-box is made using a spring of natural length 0.2m and modulus 100N and a 'jack' of mass 0.5 kg. When the lid is closed the spring is compressed to a length of 0.1m. Assuming that the spring is vertical throughout, calculate the maximum distance that the 'jack' will rise when the lid is suddenly raised.



The answer is 0.180m but I can't seem to get that answer!!!!!

When lid is closed
EPE = 100(0.1)^2/2(0.2) = 2.5J
GPE = 0.5g(0.1) = 0.49J
KE = 0

At max distance
EPE = 100x^2/2(0.2) = 250x^2
GPE = 0.5g(0.2+x) = 0.98 + 4.9x
KE = 0

=> 2.5 + 0.49 = 250x^2 + 0.98 + 4.9x
250x^2 + 4.9x - 2.01 = 0

x = [-4.9 +/- rt(4.9^2 - 4(250)(-2.01))]/500
x = 0.0804 or x = -0.1 (initially)

Hence, 'jack' rises a distance = 0.1m + 0.0804m = 0.180m (3sf)
Reply 2
my solution to Q3 is always wrong, its depressing me. (same exercise)

any help will be very much appreciated
Reply 3
d'oh! i can't do that question either. i hate this part of elasticity! i can never get the energies right.. *wails*
Reply 4
I'm just getting stuck on all of them.
A particle P of mass 0.5kg is attached to one end A of a light elastic string of natural length 2m and modulus 20N. The other end B of the string is fixed to a point on the ceiling. The particle is held at a distance of 1.5m vertically below B and then released. Calculate (a) the length of the string when the particle reaches its lowest point (b)the speed of the particle when it passes through its equilibrium position.

a)
Energy at 1.5m below B
GPE = -0.5g(1.5) = -7.35
EPE = 0
KE = 0

Energy at lowest point
GPE = -0.5g(2+x) = -9.8 - 4.9x
EPE = 20x^2/2(2) = 5x^2
KE = 0

Energy at 1.5m below B = Energy at lowest point
-7.35 = -9.8 - 4.9x + 5x^2
5x^2 - 4.9x - 2.45 = 0

x = [-(-4.9) +/- rt((-4.9)^2-4(5)(-2.45))]/2(5)
x = [4.9 +/- rt(73.01)]/10
x = 1.34m or x = -0.364
x>0 => length of string at lowest point = 2m + 1.34m = 3.34m

b)
Equlibrium => Tension = weight of P
=> 20e/2 = 0.5g
e = g/20m = 0.49m

At x = 0.49m
KE = 0.5(0.5)v^2 = 0.25v^2
GPE = -0.5g(2+0.49) = -12.201
EPE = 20(0.49)^2/2(2) = 1.2005

Energy at (x = e = 0.49m) = Energy at 1.5m below B
=> 0.25v^2 - 12.201 + 1.2005 = -7.35
=> 0.25v^2 = 3.6505
v^2 = 14.602
v = 3.82m/s
Reply 6
Yup, managed to do it last night, after deliberating for quite a while :smile:

Widowmaker, do you have any tips for these kinds of questions? I know you need to consider all the energies, and that Initial total energy = Final total energy, but when it comes to forming the equations I become lost!

Thanks again
Shickles
Yup, managed to do it last night, after deliberating for quite a while :smile:

Widowmaker, do you have any tips for these kinds of questions? I know you need to consider all the energies, and that Initial total energy = Final total energy, but when it comes to forming the equations I become lost!

Thanks again

Draw a very good diagram. :wink:
Reply 8
why cant strings just be inelastic like in the good old m1 and m2 days...
Original post by benji12394
why cant strings just be inelastic like in the good old m1 and m2 days...


You resurrected this thread just to make that comment?! FS!
(edited 13 years ago)