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InterviewSolution
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A man uses a rough inclined plane of length 3 m to raise a load of 100 kgwt. Ifhe does 2400 J of work and the inclined plane offers 300 N resistance,find the mechanical advantage (Take g=10 m s^(-2)). |
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Answer» Solution :Work done to raise the load = Work done against gravity + work done agianst friction, w = MG ` Sin theta xx 1 + f xx 1` SubstitutingW = 2400 J, mg = 1000 N, 1=3 m, f = 300 N, we get 2400 J = `(1000xx sin theta + 300)xx3` `100 sin theta + 300 = 800` `1000 sin theta = 500` ` sin theta = (500)/(1000)` `sin theta = (1)/(2)` M.A. ` = ("load")/("effort")=(mg)/( mg sin theta + f_(s))` substitute mg` = 1000 N sin theta = (1)/(2)`. `f_(s)=300 N` `MA = (1000)/(1000xx(1)/(2)+300)=(1000)/(800)=(5)/(4)= 1.25` Alternative method : Lengthof the given inclined PLANE 1 = 3 m (given) Work done by the effort, `W_(E)=2400 J ` (given) Let 'E' be the effort. `:. W_(e)=El rArr 2400 = E (3) rArr E = 800 N` Mass of the load = 100 kgwt `:.` Load L = (100) (10) `[:' g = 10 m s^(-2)]` `= 1000 N` `:.` Mechanical advantage (M.A.) ` = (L)/(E)=(1000)/(800)=1.25` |
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