Chemistry, asked by sunitatyagi848, 6 months ago

A block A of mass m1 is released from top of smooth inclined plane and it slides down the plane. Another block of mass m2 such that m2 > m1is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?

(a) Both blocks will reach ground at same time

(b) Both blocks will reach ground with the same speed

(c) speed of both the blocks when they reach ground will depend on their masses

(d) Block A reaches ground before block B

Answers

Answered by gautamjainvtpl

Answer:

$i \: think \: it \: must \: be \: option \: c$

Answered by Rameshjangid

Answer:

Both the blocks will reach to the ground with the same speed,because the potential energies of both decrease by the same amount, which gets converted into kinetic energy, option (a) is correct.

Explanation:

When the block is released, it moves down the plane under a force mg sinθ.Hence the acceleration of the block down the plane is

a = g sinθπ

If the block starts from rest from point A, then its velocity when it reaches the bottom B is given by

$v^{2}-u^{2} = 2as \\v^{2}-0 = 2aL$

$v=\sqrt{2aL}=\sqrt{2g(Lsinθ)}=\sqrt{2gh}\\$

Here, $θ$ = θ

For the block which falls vertically downward in height,

$v^{'}= \sqrt{2gh}$

So, the speed of both blocks will be the same.