Q.

The masses of the blocks A and B are 0.5 kg and 1 kg respectively. These are arranged as shown in the figure and are connected by a massless string. The coefficient of friction between all contact surfaces is 0.4. The force ( in N ) necessary to move the block B with constant velocity will be (g=10 m/s2)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

answer is 10.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here, mA=0.5 kg;mB=1 kg Force on block A  T=μmAg..............(1)  Force acting on block B F=T+μmAg+μmA+mBg...........(2)  From (1)&(2) F=μmAg+μmAg+μmAg+μmBg F=3μmAg+μmBg=μg3 mA+mB =0.4×10×(3×0.5+1)=10 N
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon