Conservation of energy problem

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Homework Statement


An elastic string of natural length 3.0m can be stretched to a length 4.0m by a weight of mass 10.0kg. Its two extremities are fixed to two points A, B in the same horizontal line at a distance of 4.0m apart, and a mass of 15.0kg is attached to the mid-point. If this mass is released from rest while the string is horizontal, find the velocity of the mass when it has descended a distance of 1.5m


Homework Equations





The Attempt at a Solution


F = kx
10g = k(1)
k = 10g

15g = 10gx
x = 1.5m (Maximum stretched length of the string)

When the mass has fallen a distance of 1.5m, the string is stretched to the length 2(1.52+22)1/2= 5

Thus, stretched length of the string = 1m

0.5(10g)(1.5)2=0.5(10g)(1)2+0.5(15)(v2)
v = 2.86m/s

However, the answer is 3.13m/s. Where did I go wrong?
 

Answers and Replies

  • #2
try to do it by energy consideration. the natural length is 3m and it has been tied between points 4m distant. so the string is already stretched. the energy stored in the string is 1/2 kx2. the mass is initailly 1.5m higher than now(i,e, when it has fallen 1.5m). so the initial energy of the system is 1/2kx2 + mgh. the final energy is 1/2ky2 + 1/2mv2, where y is the current ellongation of the string and v is to be determined. find y by using pythagorian theorem using h and the distance between tied end and u will have only one unknown, in v, to be solved out.
 
  • #3
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When the mass has fallen a distance of 1.5m, the string is stretched to the length 2(1.52+22)1/2= 5
Good.

Thus, stretched length of the string = 1m
What's the unstretched length of the string? In its initial position, how much is it stretched? In its final position?

Don't neglect gravitational PE when analyzing energy.
 

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