Question

5.) 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 ) (1) 2000 j (2) 4000 j (3) 6000 j (4) 8000 j

Answer 1

The work done on the man by the floor will be equal to the sum of change in potential energy and kinetic energy.
i-e, W = Δ P.E + Δ K.E 
      W = mgh + 1/2 mv²
      W = m (gh + v²/2)   ............ (1)
To solve this, first we have to calculate the value of "v"
Using equation of motion:
Vf² - Vi² = 2aS
If elevator starts its motion from rest then its velocity at distance 12 m is
v² - 0 = 2 x 10/3 x 12
v² = 80
Now put this value in equation (1), we get:
W = 50 [(10).(12) + 80/2]
W = 50 (120 + 40)
W = 50 x 160
W = 8000 J
which shows that option (4) is correct.
Hope it will help you.

Answer 2

Explanation:

here the man is standing in the elevator so there must be a kinetic engery and potential energy also . bln refer the sum