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A block of mass 1.5 kg is allowed to slide down along a quadrant of a circle from the horizontal position in reaching to the bottom its velocity is 8 M per second the work done is overcoming the friction is 12j the radius of circle is
Using the conservation of energymgh=1/2 mv²+ 12given mass =1.5Kgvelocity=8m/s1.5*10*h=1/2*1.5*8²+1215h=60h=4mso the radius also should be equal to 4m
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