Question

A block is placed on the top of a plane inclined at ${37}^{\circ }$ with horizontal. The length of the plane is 5m. The block slides down the plane and reaches the bottom. Find the speed of the block at the bottom if the coefficient of friction is 0.25.

• A. $\sqrt{2g}$
• B. $\sqrt{4g}$
• C. $\sqrt{g}$
• D. None of these

Answer 1

The correct option is B

$\sqrt{4g}$

Point out the forces on the block in its motion.

(1) mg; downwards

(2) Normal; perpendicular to plane

(3) Friction; opposite to direction of motion.

We know $W_{net}=\Delta K.E$.

$W_{mg}+W_N+W_{fr}=\Delta K.E$.
$\left(mgsin\theta \right)l-\left(\mu mgcos\theta \right)l=\frac{1}{2}m{v}^{2}mgl[sin\theta -\mu cos\theta ]=\frac{m{v}^2}{2}v=\sqrt{2gl(sin\theta -\mu cos\theta )}=\sqrt{2\times g\times 5(\frac{3}{5}-\frac{1}{4}\times \frac{4}{5})}=\sqrt{4g}$