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A disc of mass 100 g slides down from rest on an inclined plane of 30^(@) and come to rest after travelling a distance of 1m along the horizontal plane. If the coefficient of friction is 0.2 for both inclined and horizontal planes, then the work done by the frictional force over the whole journey, approximately, is (Acceleration due to gravity, g = 10 ms^(-1) ) |
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Answer» 0.106 J From third equation of the motion, velocity of disc when it leaves the inclined plane,  `V^(2) - u^(2)- 2as` or `"" mu = sqrt(2as)""(because v = 0)` `therefore` Acceleration of a block on a horizontal, a = `mu` G `rArr"" a = 0.2 xx 10 = 2m//s^(2)"" ("given", mu = 0.2 )` After putting the value of a in Eq. (i), we get `therefore "" u = sqrt(2 xx 2 xx 1 )` = 2 m/s `"" ( because s = 1 ` m, given ) From above diagram, The FRICTIONAL force applied on the disc inclined plane r is F = `mu` N = `mu` mg cos `30^(@)` and the net acceleration force down the inclinedplane, mg sin `30^(@)` -f = mg sin `30^(@) - mu mg cos 30^(@) = ma_(1)` or `a_(1) = g(sin 30^(@) = mu cos 30^(@)) = 5( 1 - sqrt(3) mu )`... (1) HENCE, from the third equation of motion, `v_(1)^(2)= mu_(1)^(2) + 2a_(1) `s When s is distance travelled by the disc, `v_(1) = u = 2 m//s m_(1) = 0` `therefore "" (2)^(2) = 2a_(1) s " or " s = (2)/(a_(1)) ` Putting the value of `a_(1)` form Eqs (i) we get `rarr "" s = (2)/(5 (1 - sqrt(3) mu ) ) ` = `(2)/( 5 (1 - sqrt(3) xx 0.2)) = 0.612 `m Hence, the work done by the frictional force, W = work done on inclined plane + work done on horizontal plane. = `( mu mg cos 30^(@) ) s + (1)/(2) mu^(2)` Putting the given values , we get ` = 0.2xx (100)/(1000) xx 10 xx cos 30^(@) xx 0.612 + (1)/(2) xx (100)/(1000) xx (2)^(2)` = 0.306 J |
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