Question 110: Laws of Motion

Question

Consider a block kept on an inclined plane (incline at 45$$^\circ$$) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane($$\mu$$) is equal to :

Accepted Answer

$$
\begin{aligned}
& \mathrm{F}_1=\mathrm{mg} \sin 45^{\circ}+\mathrm{f}=\mathrm{mg} \sin 45^{\circ}+\mu \mathrm{N} \\
& \mathrm{F}_1=\frac{\mathrm{mg}}{\sqrt{2}}+\mu \mathrm{mg} \cos 45^{\circ} \\
& \mathrm{F}_1=\frac{\mathrm{mg}}{\sqrt{2}}(1+\mu)
\end{aligned}
$$
$$
\begin{aligned}
& F_2=m g \sin 45^{\circ}-f=m g \sin 45^{\circ}-\mu N \\
& =\frac{m g}{\sqrt{2}}(1-\mu)
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{F}_1=2 \mathrm{~F}_2 \\
& \frac{\mathrm{mg}}{\sqrt{2}}(1+\mu)=2 \frac{\mathrm{mg}}{\sqrt{2}}(1-\mu) \\
& 1+\mu=2-2 \mu \\
& \mu=1 / 3=0.33
\end{aligned}
$$

Answer

0.60

Answer

0.33

Answer

0.25

Answer

0.50

Laws of Motion

Consider a block kept on an inclined plane (incline at 45$^\circ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane(\mu$) is equal to :

Choose the correct answer: