Question
A block of metal of mass 50 g when placed over an inclined plane at an angle of $15^{\circ}$ slides down without acceleration. If the inclination is increased by $15^{\circ}$, what would be the acceleration of the block?
✓
Answer
Here $m =50 g=0.05 kg$
Angle of repose,
$
\begin{aligned}
& \alpha=15^{\circ} \\
& \therefore \mu=\tan \mu=\tan 15^{\circ}=0.2679
\end{aligned}
$
New angle of inclination $=15+15=30^{\circ}$
Let a be the downward acceleration produced in the block.
Net downward force on the block is
$
F=mg \sin \theta-f
$
$\begin{aligned} & ma = mg \sin \theta- u mg \cos \theta[\because f=\mu R=\mu m g \cos \theta] \\ & \therefore \quad a=g(\sin \theta-\mu \cos \theta) \\ & =9.8\left(\sin 30^{\circ}-0.2679 \cos 30^{\circ}\right) \\ & =9.8(0.5-0.2679 \times 0.866) \\ & =9.8 \times 0.2680=2.6 ms^{-2}\end{aligned}$
Angle of repose,
$
\begin{aligned}
& \alpha=15^{\circ} \\
& \therefore \mu=\tan \mu=\tan 15^{\circ}=0.2679
\end{aligned}
$
New angle of inclination $=15+15=30^{\circ}$
Let a be the downward acceleration produced in the block.
Net downward force on the block is
$
F=mg \sin \theta-f
$
$\begin{aligned} & ma = mg \sin \theta- u mg \cos \theta[\because f=\mu R=\mu m g \cos \theta] \\ & \therefore \quad a=g(\sin \theta-\mu \cos \theta) \\ & =9.8\left(\sin 30^{\circ}-0.2679 \cos 30^{\circ}\right) \\ & =9.8(0.5-0.2679 \times 0.866) \\ & =9.8 \times 0.2680=2.6 ms^{-2}\end{aligned}$
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