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JEE Advanced Model Paper 17 with solutions for online practice
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© examsnet.com
Question : 20
Total: 60
An atwood machine is setup in an elevator moving upward at 5m/s and slowing down at 2 m/
s
2
. The initial velocity of block B is 2 m/s upward and the acceleration of block A is 3 m/
s
2
downwards. Find the time (in sec.) at which block B will return to its initial position.Assume the string remains taut and the acceleration of the elevator does not change duringthe required time interval.
Your Answer:
Validate
Solution:
Acceleration of B let a
a + 3 = 2 × 2
a = 1 m/
s
2
So s = 2 × t -
1
2
× 1 ×
t
2
= 0
t = 4 sec
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