Question

A block is placed on the top of a plane inclined at ${30}^{\circ }$ with horizontal. The length of the plane is $10\text{m}$. The block slides down the plane and reaches the bottom. If the surface is frictionless, then the speed of the block at the bottom of the inclined plane is

• A. $10\text{m/s}$
• B. $10\sqrt{2}\text{m/s}$
• C. $5\text{m/s}$
• D. $5\sqrt{2}\text{m/s}$

Answer 1

The correct option is A $10\text{m/s}$

Given, length of inclined plane $\left(l\right)=10\text{m}$
Height of inclined plane $\left(h\right)=10\sin {30}^{\circ }=5\text{m}$
As the block slides down the inclined plane, it loses potential energy and gains kinetic energy
Loss in P.E. = Gain in K.E. (Law of conservation of energy)
$mgh=\frac{1}{2}m{v}^2$
$v=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10\text{m/s}$