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Q)A block of mass m is attached with a massless spring of force constant k.The block is placed over a rough inclined surface for which coefficient of friction is ¾.Find the minimum value of M required to movethe block up the plane.(neglect friction in pulley.neglect mass of string and pulley).See the figure in the attachment.plz solve and explain sir.

Sibashis ghosh , 9 Years ago
Grade 11
anser 3 Answers
Aakanksha Talreja

Last Activity: 9 Years ago

Forces on block of mass m – normal of incline,mgsin37,mgcos37,\muN(in direction of mgsin37 as block is resting),kx(in opposite direction of mgsin37).
Forces on block of mass M – tension T and Mg downward
As block is placed over inclined i.e. it is in rest.Hence, T=Mg , T=kx and T=mg\sin 37 + \mumg\cos 37=6mg/5
\therefore M=6m/5.

Yuvrajpratapsingh

Last Activity: 8 Years ago

This can be solved in a better and quicker method using work. Let spring be displaced by a distance x. So M moves by a distance x too. 1/2kx^2=Mgx......equation 1Now as block moves up friction is applied downwards and kx up the plane, Kx=mgsin(theta)+3/4mgcos(theta)Kx=6mg/5Replacing this in equation 1We have M=3m/5

Divyansh

Last Activity: 6 Years ago

f+mgsin(theta)=tensionf=friction ....f=3/4mgcos(theta)tension=Mgequate it and yu will get the answer

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