IIT-JEE questions.
Classical Mechanics
Level
2
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstreched. Then the maximum extension in the spring is
Mg/2k
4 Mg/k
2 Mg/k
Mg/k
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
for maximum extension work done = 0 (as initial K.E = final K.E ) . Hence Mgx - 1/2(k*xsquare) = 0. so value of x( maximum extension) comes out to be = 2Mg/k