17. A body is released from position A as shown in figure. The speed of body at position B is
[NCERT Pg. 121]
22
50 m
- В
10 m
]
(1) 10 m/s
(2) 10/2 m/s
(3) 20 m/s
(4) 2012 m/s​

Potential energy at A is mgh1 = 50mg

Energies at point B are kinetic and potential.

Potential energy at B is mgh2 = 10mg

Kinetic energy at B is 1/2mv^2

Energy at A = Energy at B

mgh1 = mgh2 + 1/2mv^2

50mg = 10mg + 1/2mv^2

1/2mv^2 = 40mg

1/2v^2 = 40g

v^2 = 40 x 10 x 2

v^2 = 800

v =✓800 m/s

v = 28.284

HOPE IT HELPS!!!

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.