Question 4: Laws of Motion

Question

A piece of ice slides down a rough inclined plane at $$45^{\circ}$$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

Accepted Answer

Given, $$	heta=45^{\circ}, s_1=s_2, u=0$$
On the rough incline, $$a_1=g(\sin 	heta-\mu \cos 	heta)$$
$$t_1=$$ time taken
On the frictionless incline, $$a_2=g \sin 	heta$$
$$\begin{aligned}
& t_2=	ext { time taken, } t_1=2 t_2 \
& s=u t+\frac{1}{2} a t^2 \
& s_1=0+\frac{1}{2} g(\sin 	heta-\mu \cos 	heta) t_1^2 \
& s_2=0+\frac{1}{2} g \sin 	heta t_2^2
\end{aligned}$$
$$	ext { As } s_1=s_2 	ext {, }$$
$$\begin{aligned}
& \frac{1}{2} g(\sin 	heta-\mu \cos 	heta) t_1^2=\frac{1}{2} g \sin 	heta t_2^2 \
& \frac{\sin 	heta-\mu \cos 	heta}{\sin 	heta}=\frac{t_2^2}{t_1^2} \
& \Rightarrow \quad 1-\mu \cot 	heta=\frac{t_2^2}{\left(2 t_2ight)^2} \
& \Rightarrow \quad 1-\mu \cot 	heta=\frac{1}{4} \
& \Rightarrow \quad \mu \cot 	heta=1-\frac{1}{4}=\frac{3}{4} \
& 	herefore \quad \mu=\frac{3}{4 \cot 	heta} \
&
\end{aligned}$$

Answer

$$\frac{3}{7 \cot 	heta}$$

Answer

$$\frac{4}{7 \cot 	heta}$$

Answer

$$\frac{3}{4 \cot 	heta}$$

Answer

$$\frac{7}{9 \cot 	heta}$$

Laws of Motion

A piece of ice slides down a rough inclined plane at $45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

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