CBSE Class 11-science Answered
Two blocks of mass m=5kg and M=10 kg are connected by a string passing over a pulley B as shown. Another string connects the centre of the pulley B to the floor and passes over another pulley A as shown. An upward force F is applied at the centre of pulley A. Both the pulleys are massless. Find the acceleration of both the blocks if 1) F=100N. 2)F=500N
![question image](https://images.topperlearning.com/topper/new-ate/639597964da0936bf8dfLawsofMotionDPP1Q1.png)
Asked by sakethrockzz007 | 14 Aug, 2023, 16:30: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/1cca6b270a77c00ec142b589338ecd1d64da607ce37282.56596250f4.png)
Figure shows the free body diagrams of pulleys and blocks. F is applied force on pulley-A .
TA and TB are tension forces along the cables connected to pulley.
(mg) is weight of block of mass m . (Mg) is weight of block of mass M.
In pulley-A , since pulley's mass is negligible , Tension forces TA balances the applied force.
Hence , we have,
![begin mathsize 14px style F space equals space 2 space T subscript A end style](https://images.topperlearning.com/topper/tinymce/cache/95ebcbfc9ac2531079fd7bf9d0b80900.png)
![begin mathsize 14px style T subscript A space equals space F over 2 end style](https://images.topperlearning.com/topper/tinymce/cache/e58ebcf7d9d6d19a27b33389dae99331.png)
In pulley-B , we have
![begin mathsize 14px style 2 space T subscript B space equals space T subscript A end style](https://images.topperlearning.com/topper/tinymce/cache/dbfb7c93ce433f5c1345a2cfa5404e0d.png)
![begin mathsize 14px style T subscript B space equals space 1 half space T subscript A space end subscript equals space 1 half cross times F over 2 space equals space F over 4 end style](https://images.topperlearning.com/topper/tinymce/cache/9753c757debe8dc8f714518a853f56a2.png)
If applied force F = 100 N , then TB = 25 N , block of mass 5 kg moves downward with acceleration am
![begin mathsize 14px style m g space minus space T subscript B space equals space m space a subscript m end style](https://images.topperlearning.com/topper/tinymce/cache/384f28514d34f414ac7d342bd722ef85.png)
![begin mathsize 14px style a subscript m space equals space g space minus space T subscript B over m end style](https://images.topperlearning.com/topper/tinymce/cache/51662f3d26de46bcf09e68aba6e18584.png)
![begin mathsize 14px style a subscript m space equals space left parenthesis space 9.8 space minus space 25 over 5 right parenthesis space m divided by s squared space equals space 3.8 space m divided by s squared end style](https://images.topperlearning.com/topper/tinymce/cache/8dd1e3baeb32b1f013b486f62df064a6.png)
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Let aM be the acceleration of block of mass M
![begin mathsize 14px style M space g space minus space T subscript B space equals space M space a subscript M end style](https://images.topperlearning.com/topper/tinymce/cache/7e9912a1ad7d2eab203216337d7a70bf.png)
Hence acceleration aM is
![begin mathsize 14px style a subscript M space equals space g space minus space T subscript B over M end style](https://images.topperlearning.com/topper/tinymce/cache/6f681d085d806c0e3fc082f47f788aae.png)
![begin mathsize 14px style a subscript M space equals space left parenthesis space 9.8 space minus space 25 over 10 right parenthesis space m divided by s squared space equals space 7.3 space m divided by s squared end style](https://images.topperlearning.com/topper/tinymce/cache/0a5161229a333a836a4b06d639133d42.png)
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If applied force F = 500 N , then TB = 125 N .
Block of mass 5 kg moves upward with acceleration am .
![begin mathsize 14px style m space a subscript m space equals space T subscript B space minus space m g space end style](https://images.topperlearning.com/topper/tinymce/cache/8a91ca2b93f96a57808d272edb9dafab.png)
![begin mathsize 14px style a subscript m space equals space T subscript B over m minus space g end style](https://images.topperlearning.com/topper/tinymce/cache/398093708f73538c6ceba327418f3803.png)
![begin mathsize 14px style m space equals space left parenthesis space 125 over 5 space minus space 9.8 space right parenthesis space m divided by s squared equals 15.2 space m divided by s squared space end style](https://images.topperlearning.com/topper/tinymce/cache/264e5103e591c3a7daa00254d4b3bb6a.png)
--------------------------
If aM is the acceleration of block of mass M , then we have
![begin mathsize 14px style M space a subscript M space equals space T subscript B space minus space M g end style](https://images.topperlearning.com/topper/tinymce/cache/99a0096ee56937056c3ece0c4692ba9f.png)
![begin mathsize 14px style a subscript M space equals space T subscript B over M minus space g space equals space left parenthesis space 125 over 10 minus 9.8 space right parenthesis space m divided by s to the power of 2 to the power of space end exponent space equals 2.7 space m divided by s squared space end style](https://images.topperlearning.com/topper/tinymce/cache/c9f857c75bef27b6ccb246c7ca604b77.png)
Answered by Thiyagarajan K | 14 Aug, 2023, 23:26: PM
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