CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

Block of mass m is released from rest when spring is in its natural length. Assume pulley is ideal and block does not strike on ground during its motion in vertical plane. Then:


A
Maximum elongation in spring is 4mgk
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Maximum elongation in spring is 2mgk
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Maximum speed of block is 2gmk
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Maximum speed of block is gmk
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C Maximum speed of block is 2gmk
When block is released, the spring will stretch by x and block will move downwards by 2x.

By energy conservation principle,

Energy gain by spring = Loss in gravitational potential energy by block
12kx2max=mg(2xmax)

xmax=4mgk ( at maximum elongation)

The block will attain maximum velocity when net force on the pulley is zero.
Let net force on block be zero when x=x

At equilibrium, kx=2mg

x=2mgk

The kinetic energy of the block is given by
12mv2=mg(2x)12kx2

12mv2max=mg(2x)12kx2

Substituting x to get maximum velocity (vmax)

12mv2max=2mg(2mgk)12k(2mgk)2
12mv2max=2m2g2k
vmax=2gmk

Hence, options (a) and (c) are the correct answers.

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image