10 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.like i know initially speed will be zero but then will it not attain a specific velocity to complete quarter circle ie root 2gr
and like if are applying it wouldnt final U = 0 cuz the moment it is released it is still there at the same height
the potential energy of spring will convert to kinetic energy and at that point block will have root2gr velocity then at peak that kinetic energy will be 0 and gravitational potential mg2R
are you asking what if we take potential at top to be our reference ?
It is given that "the block pushes with a force of mg", i.e., normal at point P is mg=mv^2/r. So the K.E.=mgr/2. Also, the P.E. at that point is mgr. So, 3mgr/2=1/2kx^2
OH HE MENT THAT BY THE :Saul_Goodman:
QUESTION
oh
its gna take me a minute to decipher all this ðŸ˜
x^2 = 3mgr/k ??
okiiiie got it now
she but idc
thanks
+solved @hardcoreisdead @727 @SirLancelotDuLac @
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