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A block of mass 25 kg …

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A block of mass 25 kg is lying on a rough horizontal surface . When a force of 122.5 newton is applied horizontal surface, the block just begins to move . Find the coefficient of friction between the surface and the block

Posted by Anajli Gariya 6 years, 2 months ago

CBSE > Class 11 > Physics

3 answers

Anajli Gariya 6 years, 2 months ago

I thi

Rohan Singh 6 years, 2 months ago

M=25kg R=m×g =250N Fsmsx=122.5N so, Fsmax=u×R 122.5N=u×250N =. u=49 Cofficient of friction=49

Beast Boy 6 years, 2 months ago

Hi there..? We know that Frictional force= coefficient of friction(¶) × Resultant force(R) F=¶×R F/R=¶ Now, resltuant force =mass × g R=25×10 R=250 And it is given that F=122.5 N Puting these values in the equation formed 122.5/250=¶ We get 0.5 So this is the ans If u liked it plz follow me on Instagram lesner321 And happy diwali in advance...???✨?

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