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

A block of mass M is placed on the top of a wedge of mass 4M. All the surfaces are frictionless. The system is released from rest. The distance moved by the wedge at the instant the block reaches the bottom will be :
3557_4d3d6905b7b54c109c41ca235e70957e.png

A
0.2m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.4m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.8m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.8m
If x be the distance moved by the wedge of mass 4M at instant

when block M reaches at the bottom, then block M also move

(4x) distance at that instant.
Therefore, 4Mx=M(4x) (as work done by block M is equal to the

work done by wedge of mass 4M)
5Mx=4M
x=45=0.8 m

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
What's with All the Tension in the Room?
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon