wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

If block A is moving horizontally with velocity vA, then find the velocity of block B at the instant shown in figure.


A
hvA2x2+h2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
xvAx2+h2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
xvA2x2+h2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
hvAx2+h2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C xvA2x2+h2

Let the vertical distance of block B from the ground be y.
Hence, total length of string L is:
L=2h2y+x2+h2
Differentiating the equation w.r.t time, we get:
0=02dydt+12x2+h2×2x×dxdt
(since h is constant)
dydt=x2x2+h2dxdt ....(i)
But dydt=vB & dxdt=vA ....(ii)
because vB= time rate of change of vertical distance y between ground and block B
vA= time rate of change of horizontal distance x between block A and block B

From Eq. (i) and Eq. (ii):
vB=xvA2x2+h2
Hence option (c) is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon