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A block of mass m is released on the top of a smooth inclined plane of length x and inclination theta. Horizontal surface is rough. If block comes to rest after moving a distance d on the horizontal surface, then coefficient of friction between block and surface will be

SARDESAI
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F=mu N
therefore 
mu=F/N
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Since inclined surface is smooth and horizontal surface is rough thus, mgh = @mgd ........ ( @ -> mu ) But, H = xsin(theta) Thus, mgxsin(theta) = @mgd or, xsin(theta) = @d or, @ = xsin(theta)/d
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Work energy theorem u=0 v=0 Ki=0 Kf=0 ☆=meu mgh-☆mgd=0 ☆=h/d h=xsin theta ☆=x sin theta/d
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a block

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By work energy theorem

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solve this question mentioned above
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