A Man Of Mass 50 Kg Is Standing In An Elevator. If Elevator Is Moving Up With An Acceleration G 3 Then Work Done By Normal Reaction Of Floor Of Elevator On Man When Elevator Moves By A Distance 12 M Is (G= 10 M/S 2)(A) 8000 J, (B) 6000 J, (C) 4000 J,(D) 2000 J

The correct answer is (A) 8000 J

Sol:

The work done on the man by the floor will equal his change in potential energy plus his change in kinetic energy.

If we assume the elevator starts at rest, the velocity at the distance of 12 m can be found using the kinematic equation.

[latex]V^{2} = V_{0}^{2} + 2ad [/latex] [latex]\Rightarrow V^{2} = 0^{2} + 2(10/3)12 [/latex] [latex]\Rightarrow V^{2} = 80 [/latex] [latex]U = \Delta PE + KE [/latex] [latex]\Rightarrow U = mgh + 1/2 mv^{2} [/latex] [latex]\Rightarrow U = m (gh + 1/2 v^{2}) [/latex] [latex]\Rightarrow U = 50 ((10)(12) + 1/2 80) [/latex] [latex]\Rightarrow U = 50 ((10)(12) + 1/2 80) [/latex] [latex]\Rightarrow U = 50 (120 + 40) [/latex] [latex]\Rightarrow U = 50 (160) [/latex] [latex]\Rightarrow U = 8000 J [/latex]

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