CBSE Class 11-science Answered
a block of mass m is pulled along a horizontal smooth surface by a rope of mass m force p applied at one end of rope force which the rope exert on block is
![question image](http://images.topperlearning.com/topper/new-ate/top_mob169992291537583585470398482-d132-4bed-ba15-cf2e0e2fb561.jpg)
Asked by rame123shshiv | 14 Nov, 2023, 06:18: AM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/b4e1bcaa7741d46656586c57dca9c1df6552f3f8979fc0.09318199f6.png)
Fig.1 shows the system of mass m and rope of equal mass m is pulled by a force P .
The system is under accelerated motion with acceleration a due to the applied force P.
Fig.2 shows the free body diagram of rope and mass m showing the forces acting on rope and mass m .
F is the contact force between mass m and rope.
By Newton's third law , contact forces F acting on rope and mass are action-reaction pair ,
equal in magnitude and opposite in direction.
If we apply Newton's second law to rope , we get
![begin mathsize 14px style P space minus space F space equals space m space a end style](https://images.topperlearning.com/topper/tinymce/cache/4f512e71e1b26eb188762f992fced5b6.png)
If we apply Newton's second law to mass , we get
![begin mathsize 14px style F space equals space m space a end style](https://images.topperlearning.com/topper/tinymce/cache/369ca83fb837db44ec940aef08a71ce7.png)
By adding eqn.(1) and (2) , we get , P = ( 2 m ) a
acceleration a = P / (2m)
Hence, we get contact force F from eqn.(2) , we get
F = P/2
Hence force exerted by rope on mass m is P/2
Answered by Thiyagarajan K | 14 Nov, 2023, 10:07: AM
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