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

A block A (5 kg) rests over another block B (3 kg) placed over a smooth horizontal surface. There is friction between A and B. A horizontal force F1 gradually increasing from zero to a maximum is applied to A so that the blocks move together without relative motion. Instead of this another horizontal force F2, gradually increasing from zero to a maximum is applied to B so that the blocks move together without relative motion. Then

A
F1 (max)=F2 (max)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
F1 (max)>F2 (max)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
F1 (max)<F2 (max)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
F1 (max):F2 (max)=5:3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D F1 (max):F2 (max)=5:3
Case I:
Since, no relative motion, acceleration of both the blocks will be equal
a=F1Ff5=Ff3F1(max)=83Ff
Case II:
Again, since no relative motion, acceleration of both the blocks will be equal
a=Ff5=F2Ff3F2(max)=85Ff
Clearly;
F1 (max)>F2 (max) and F1 (max)F2 (max)=53

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Warming Up: Playing with Friction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon