Physics Problem Solving (Q -15), IIT-JEE physics classes - Videos


Watch Solution here..
Q.15- When a 0.26-kg block is suspended on a vertical spring, it causes it to stretch 2.50 cm. If the block is now pulled 8.40 cm below its equilibrium position and released, what is the speed of the block when it is 4.30 cm below the equilibrium position?
*Don’t forget to hit SHARE, LIKE and SUBSCRIBE!
*SUBSCRIBE( it’s free):
Our motto is “Education is Free and for all.”
* Watch One Video Daily as Physics Tonic to Ensure your Success in IIT- JEE exams with Higher Ranks. Start now , Copy Paste this link in the browser:-
IIT-JEE Physics Classes Playlist
*Physics Full Playlists:-
Error & Uncertainties, Significant Figures:
Vernier Caliper ( How to read Measurements):
Screw Gauge : (Least Count, Reading Measurements):
Dimensional Analysis:
Kinematics: Position-Time Graph, Velocity:
ap physics lectures,
ap physics
ap physics 1
ap physics c
ap Physics c mechanics
physics 1
ap physics 2
ap physics b
ap physics study guide
ap physics help
ap physics review
SAT / ACT Aptitude Tests



Please enter your comment!
Please enter your name here