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Grade: 11

                        

I am trying to get extension of a spring if a weight m is hanging from it. . 1. From newtons second law T-mg= 0 T = mg, kx = mg, x = mg/k 2. If I apply work energ theorem initial velocity and final velocity is zero (1/2)kx^2-mgx = 0 x = 2 mg/k Why my second method is not working to finding x, where is the mistake.

4 years ago

Answers : (1)

yash
25 Points
							
That is because both the equation though being right are telling 2 different information . The first is the x which tell you where the force becomes equal to mg that is T= mg but second tells you the distance where potential energy stored equals to mgh . Now in books we use ideal condtion saying F=-kx but in reality it means we must bring the block or spring down infitesimally slow , but in truth if you drop a object like that it will pass mg/k distance because kinetic energy is not taken into conidertation , that is the distance travelled will be 2times of the distance in yours case. In books if they ask distance moved than we use first equation but if it asks maximum distance for max energy stored or somthing like that second one.. hope you understand 
4 years ago
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