Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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.
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
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -