Question

A chain of length l and mass m is placed on a smooth surface

Answer 1

The velocity of the chain when its end reaches B is v = √  g sinθ / L  (L^2  − b^ 2 )

Explanation:

Correct statement:

A chain of length LL and mass mm is placed upon a smooth surface. The length of BA is (L-b)(L−b). Calculate the velocity of the chain when its end reaches B.

Solution:

As we know that;

Loss in P.E = gain in K.E

mg(   L/2  sinθ) - (   m/L  b) g (   b/2 sinθ)

=   1/2  mv^2  

v = √  g sinθ / L  (L^2  − b^ 2  )

​Thus the velocity of the chain when its end reaches B is v = √  g sinθ / L  (L^2  − b^ 2  )

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Answer 2

A chain of length L and mass m is placed upon a smooth surface. The length of BA is (L-b)(L−b). The velocity of the chain when its end reaches B needs to be calculated. The velocity of the chain when its end reaches B is:

v = √g sinθ / L* (L^2  − b^ 2)

1) From the laws of conservation of energy, we have:

Loss in P.E = gain in K.E

2) So, mg(L/2  sinθ) - (m/L * b) g (b/2 sinθ)=   1/2  mv^2  

Therefore, v = √  g sinθ / L  (L^2  − b^ 2)

​3) Thus the velocity of the chain when its end reaches B is :

v = √  g sinθ / L  (L^2  − b^ 2)