Energy Conservation.

  • Thread starter habibclan
  • Start date
  • #1
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Homework Statement



A 10 kg vox slides 4.0 m down the frictionless ramp shown in the following link. It then collides with the spring whose spring constant is 250 N/m.
a. What is the maximum compression of the spring?
b. At what compression of the spring does the box have its maximum velocity?

http://i196.photobucket.com/albums/aa59/aliatehreem/chapter_10.jpg


Homework Equations



Usf= Ug [ spring potential energy final = gravitational potential energy initial.
0.5 k x^2 = mgy

The Attempt at a Solution


Find height corresponding to 4 m.
h= 4 sin30

a. Then use conservation of energy to find compression.
Usf= Ug [ spring potential energy final = gravitational potential energy initial.
0.5 k x^2 = mgy
0.5 (250) (x ^2) = (10) (9.81) (4sin30)
x= 1.25 m

Apparently, the correct answer is 1.46 m for a. and 19.6 cm for b. Can someone please help me figure out what I am doing wrong? It would be greatly appreciated! I wish that they would give the length of the spring, then I could calculate the gravitational potential energy more accurately.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
The force acting on the block is mgsin(theta) and distance moved by the block is (4 + x). In solving this problem the length of the spring is not required.
 
  • #3
mgb_phys
Science Advisor
Homework Helper
7,819
14
Have you taken into account that the pe of the box is minimum when the spring stops not at first contact, so the box slides (4+x) down the slope

For (b) write the energy as KE+PE+spring=0 ( get the sign's correct! ) to find an equation only in V and x.

(Oops posts clashed)
 
  • #4
55
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by writinf 4 + x, i get 2.2 m, which is still not the correct answer.

Also, for Ke +PE + spring=0, why is the sum 0?
 
  • #5
rl.bhat
Homework Helper
4,433
8
0.5 (250) (x ^2) = (10) (9.81) (sin30)(4+x) .
Solve for x.
 
  • #6
55
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0.5 (250) (x ^2) = (10) (9.81) (sin30)(4+x) .
Solve for x.



Nice!! Thank you so much!!!!
 

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