System shown in figure is released from the rest. Pulley and string is massless and friction is absent everywhere. The speed of 5kg block when 2kg block leaves contact with ground is : (Take force constant of spring k=40N/m and g=10m/s2)
A
√2m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2√2m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4√2m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B2√2m/s Letxbetheextensioninspringwhen2kgmassleavescontactwiththeground.kx=2gx=2gk=2×1040=12mBylawofconservationofenergyfor5kgmassmgx=12k2+12mv2v=√2gx−kx2mv=√2×10×12−405×14v=2√2m/sHence,theoptionBisthecorrectanswer.