A block of mass 15kg is placed over a frictionless horizontal surface. Another block of mass 10kg is placed over it, that is connected with a light string passing over two pulleys fastened to the 15kg block. A force 80N is applied horizontally to the free end of the string. Friction coefficient between two blocks is 0.6. The portion of the string between 10kg block and the upper pulley is horizontal. Pulley, string and connecting rods are massless. (Take g=10 m/s2 )The magnitude of acceleration of the 10kg block is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
3.2 m/s2
b
2.0 m/s2
c
1.6 m/s2
d
0.8 m/s2
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
First, let us check upto what value of F, both blocks move together. Till friction becomes limiting, they will be moving together. Using the FBDs 10kg block will not slip over the 15kg block till acceleration of 15kg block becomes maximum as it is created only by friction force exerted by 10kg block on it.Let say both blocks have same acceleration.F−f10=f15 ⇒f=3F5put F=80 N⇒f=48Nfmaximum is 60 N. Therefore, for F=80 N , both will move together. Their combined acceleration, by applying NLM using both as system, F=25a⇒a=8025=3.2 m/s2