wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If block A starts from rest at t=0 & begins to move towards right with an acceleration of 2 m/s2 & simultaneously bolck C moves towards right with constant velocity of 4 m/s. Speed of block B at t=5 s will be



A
2 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 3 m/s
Velocity of block A =0+at=2t
So from 0 to 2 seconds, block A's velocity will be less than block C's velocity of 4 m/s, therefore pulley 2 and block B will move up.

At 2 seconds, velocities of block A and C are equal and therefore pulley 2 and block B will be at rest and after t=2 s, they start to move down


From the diagram, l1,l2 and l3 are the segments of the string. Therefore at t=5 s,
l1+2l2+l3=constant
Differentiating w.r.t t, we get
v1+2v2+v3=0
(l1 is decreasing and l2 and l3 are increasing)
(2×5)+2v2+4=0
2v2=6
v2=3 m/s is the velocity of block B at 5 s

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
All Strings Attached
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon