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A mass on a spring?

  • Thread starter FlukeATX
  • Start date
  • #1
7
0

Homework Statement


The figure shows an 7.9 kg stone at rest on a spring. The spring is compressed 10 cm by the stone. The stone is then pushed down an additional 32 cm and released. To what maximum height (in cm) does the stone rise from that position?


Homework Equations


k = mg/d
v = sqrt((kx^2)/m)
h = ((1/2)(v^2))/(g)



The Attempt at a Solution


So this is what I've tried:
k = mg/d
k = (7.9)(9.8) / (.1)
k = 774.2 N/m

v = sqrt((kx^2)/m)
v = sqrt((774.2*.32^2)/(7.9))
v = sqrt(79.27808/7.9)
v = 3.1678

h = (1/2v^2) / (g)
h = (1/2)(10.0352) / (9.8)
h = .512m
h = 51.2cm

Does this seem correct? I only have one shot left on the wiley plus and don't want to risk it if it's wrong. Appreciate the help!
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
I don't like the look of it! I don't think the mass will be moving when it reaches its maximum height, so no use calculating any speeds.

Seems to me you should be thinking in terms of energy.
gravitational potential energy + spring energy = constant.
mgh + 1/2k*x^2 = constant.
It shouldn't matter what reference level you use for the height, but for the extension it must be right. Would that be the initial level before the push, or before the stone is put on the spring?
 
  • #3
LowlyPion
Homework Helper
3,090
4
I would consider the total potential of when the stone was released as my starting point. (This would treat x as the distance below the relaxed spring point and y as the distance above the relaxation point. And relaxed point is where the spring would have no weight at all.)

Then you can say from conservation of energy that

1/2*k*x2 = m*g*x + m*g*y + 1/2*k*y2

You know x from 10 + 32 cm and you know k as you have found already.
 

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