1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A block of mass 10kg is attached to a relaxed string. Other end is attached to ceiling and block is released keeping the spring in vertical position. The spring is such that it has elongation of 10cm under a force of 100N. the maximum velocity of block in subsequent motion is :-

A
1m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.5m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2..0m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.5m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1m/s2First calculate spring constantP=kx100=kx100=k×0.1k=1000N/mNow calculate elongation at equilibrium positionAt equilibrium, net force F=0kx=mg1000x=10×10x=10cm=0.1mAt x=0.1m, velocity will be maximum.Using energy conservationmg×0.1=12×100×(0.1)2+12mV2max10=5+12×10V2max5=5V2maxV=1m/s2

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
The Law of Conservation of Mechanical Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program