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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(1.5

^{2}+2

^{2})

^{1/2}= 5

Thus, stretched length of the string = 1m

0.5(10g)(1.5)

^{2}=0.5(10g)(1)

^{2}+0.5(15)(v

^{2})

v = 2.86m/s

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