Question

Choose the correct answer:
1. A particle of mass m is released in a smooth
hemispherical bowl from shown position A. The
work done by gravity as it reaches the lowest
point B is (R is radius of bowl]

(1) 2mgR
(2) mgR
(3)mgR ÷ √2
(4) Zero
​

Answer 1

░▒▓█►─═ Solution. ═─◄█▓▒░

Work is defined as dot multiplication of force with displacement.

dw = f.dr.CosѲ

Point that to be noted while calculating the work done is that the direction of force applied and displacement is same.

Ѳ is the angle between the displacement and force of their respective direction are different .

And if direction is same then Ѳ = 0

And this formula reduced to

dw = f.dr

In your question....

Displacement(dr) = √(R² + R²)

Displacement = R√2

Ѳ = 45°

Force =. mg

Where m is the mass of the ball....

dw = f.dr CosѲ

put values

w = mg (R√2) Cos45°

w = mgR(√2/√2)

w= mgR

#answerwithquality

#BAL

Answer 2

Answer:

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