Question

A block of mass m is released from top of a smooth fixed inclined plane of inclination $\theta$

Find out the speed of the bIock at the bottom.

• A. $v=\sqrt{gh}$
• B. $v=\sqrt{2gh}$
• C. $v=\sqrt{4gh}$
• D. $v=\sqrt{8gh}$

Answer 1

Answer: (B)

$v=\sqrt{2gh}$

Consider u and v be the initial and final velocity of the block.

Then, by applying equations of motion we can write:

$V^2=u^2+2as$

$V^2=0+2\left(gsin\theta \right)\frac{h}{sin\theta }\Rightarrow V^2=2gh\Rightarrow V=\sqrt{2gh}$