Question

A block of metal of mass 50 gram placed over an inclined plane at an angle of 15 slides down without acceleration. If the inclination is increased by 15, what would be the acceleration of the block?
A.$2$
B. 3
C. 4
D. 5

Answer 1

Hint: On an inclined plane, the net acceleration is not zero; only the perpendicular acceleration is. This is because the normal force is equal to the gravity perpendicular component, so the net force is zero in the perpendicular direction (if not the object would go through the plane or fly off the plane upwards). Calculate friction equates it with Normal Force acting on a body and Then calculates acceleration.

Formula used:
$\mathrm{f_r}=\mu \mathrm{N}$

Complete Step-by-Step solution:
The acceleration is given by the acceleration of gravity times the sinus of the angle for a frictionless angle grade incline. "Due to an unbalanced force, objects are known to accelerate down inclined planes. The gravity force (also known as weight) acts in a downward direction; yet the normal force acts perpendicular to the surface in a direction (in fact, normal means "perpendicular").

A block of metal of mass $50 \mathrm{~g}$ when placed over an inclined plane at an
angle of $15^{\circ}$ slides down without acceleration. it means, block is moving
in equilibrium condition.
e.g., downward force along plane = upward force along plane $\mathrm{mg} \sin 15^{\circ}=\mathrm{f_r}$

we know, $\mathrm{f_r}=\mu \mathrm{N}$
and $\mathrm{N}=\mathrm{mgcos} 15^{\circ}$
so, $\operatorname{mgsin} 15^{\circ}=\mu \operatorname{mg} \cos 15^{\circ}$
$\mu=\tan 15^{\circ}$

now, when inclination increased by $15^{\circ}$
then,
 $\Rightarrow \operatorname{mgsin}\left( {{15}^{{}^\circ }}+{{15}^{{}^\circ }} \right)-\text{f_r}=\text{ma}$
$m g \sin 30^{\circ}-\mu m g \cos \left(15^{\circ}+15^{\circ}\right)=m a$
$\Rightarrow \operatorname{gsin}{{30}^{{}^\circ }}-\tan {{15}^{{}^\circ }}\cdot \text{g}\cos {{30}^{{}^\circ }}=\text{a}$
$a=g\left\{\sin 30^{\circ}-\tan 15^{\circ} \cdot \cos 30^{\circ}\right\}$
$a=10\{0.5-(2-1.73) 0.86\}$
$a=10\{0.5-0.27 \times 0.86\}$
$\therefore a=3~\text{m}/{{\text{s}}^{2}}$

The acceleration of the block be $3~\text{m}/{{\text{s}}^{2}}$. Hence, the correct option is B.

Note:
Forces are vectors and have a magnitude and a direction. The force of gravity points straight down, but it follows the ramp, but a ball rolling down a ramp doesn't go straight down. The ball is pushed into the ramp by the other component, and the ramp pushes back, so the ball does not accelerate into the ramp. It will accelerate when a rolling object, such as a ball or something on wheels, goes down a hill.