Connected bodies
In figure shown, both blocks are released from rest. The time to cross each other is
For A, 4g – T = 4a;
For B, T–1g = a ⇒ a = 3g/5
aAB = 6g/5 = 12 m/s2
⇒ 6 = 1/2 x 12 x t2 ⇒ t = 1 sec
Talk to our academic expert!
Similar Questions
A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
