A piece of ice slides down a rough inclined p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given,$	heta=45^{\circ}, s_{1}=s_{2}, u=0$.On the rough incline,the acceleration is $a_{1}=g(\sin 	heta-\mu \cos 	heta)$.On the frictionless in

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given,$	heta=45^{\circ}, s_{1}=s_{2}, u=0$.On the rough incline,the acceleration is $a_{1}=g(\sin 	heta-\mu \cos 	heta)$.On the frictionless incline,the acceleration is $a_{2}=g \sin 	heta$.Let $t_{1}$ be the time taken on the rough plane and $t_{2}$ be the time taken on the frictionless plane. Given $t_{1}=2 t_{2}$.Using the equation of motion $s=ut+\frac{1}{2}at^{2}$,for $u=0$:$s_{1}=\frac{1}{2}g(\sin 	heta-\mu \cos 	heta)t_{1}^{2}$$s_{2}=\frac{1}{2}g \sin 	heta t_{2}^{2}$Since $s_{1}=s_{2}$,we have:$\frac{1}{2}g(\sin 	heta-\mu \cos 	heta)t_{1}^{2}=\frac{1}{2}g \sin 	heta t_{2}^{2}$$\frac{\sin 	heta-\mu \cos 	heta}{\sin 	heta}=\frac{t_{2}^{2}}{t_{1}^{2}}$Substituting $t_{1}=2t_{2}$:$1-\mu \cot 	heta=\frac{t_{2}^{2}}{(2t_{2})^{2}}=\frac{1}{4}$$\mu \cot 	heta=1-\frac{1}{4}=\frac{3}{4}$$\mu=\frac{3}{4 \cot 	heta}$. Since $	heta=45^{\circ}$,$\cot 45^{\circ}=1$,so $\mu=\frac{3}{4}=0.75$.

$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ 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?

Similar Questions

$A$ block of mass $10\, kg$ is placed on an inclined plane. When the angle of inclination is $30^\circ$,the block just begins to slide down the plane. The force of static friction is ....... $kg\, wt$.

$A$ body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough,the time taken now is $2t$. The coefficient of friction of the rough surface is

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

$A$ block of mass $10 \, kg$ is moving down an inclined plane with a constant velocity of $10 \, m/s$. The angle of inclination is $37^{\circ}$. The coefficient of kinetic friction between the inclined plane and the block is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module