A smooth block is released at rest on a 45^ci | vedclass

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

Question

(A) For a smooth incline,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $d$ is $t_1 = \sqrt{\frac{2d}{g \sin 	heta}}$.Fo

Vedclass · Accepted Answer

$\mu_k = 1 - \frac{1}{n^2}$

Vedclass · Answer

(A) For a smooth incline,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $d$ is $t_1 = \sqrt{\frac{2d}{g \sin 	heta}}$.For a rough incline,the acceleration is $a_2 = g \sin 	heta - \mu_k g \cos 	heta$. The time taken to cover distance $d$ is $t_2 = \sqrt{\frac{2d}{g \sin 	heta - \mu_k g \cos 	heta}}$.Given $t_2 = n t_1$,we have $\sqrt{\frac{2d}{g \sin 	heta - \mu_k g \cos 	heta}} = n \sqrt{\frac{2d}{g \sin 	heta}}$.Squaring both sides: $\frac{1}{g \sin 	heta - \mu_k g \cos 	heta} = \frac{n^2}{g \sin 	heta}$.Since $	heta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.Thus,$\frac{1}{g \sin 	heta (1 - \mu_k)} = \frac{n^2}{g \sin 	heta}$.This simplifies to $1 - \mu_k = \frac{1}{n^2}$,which gives $\mu_k = 1 - \frac{1}{n^2}$.

$A$ smooth block is released at rest on a $45^\circ$ incline and then slides a distance $d$. The time taken to slide is $n$ times as much to slide on a rough incline than on a smooth incline. The coefficient of kinetic friction is:

Similar Questions

$A$ block of mass $m$ is at rest on an inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$,as shown in the figure. The frictional force acting on the block is:

When the angle of inclination of an inclined plane is $\theta$,an object slides down with uniform velocity. If the same object is pushed up with an initial velocity $u$ on the same inclined plane,it goes up the plane and stops at a certain distance on the plane. Thereafter,the body:

The coefficient of friction between two surfaces is $\mu = 0.8$. The tension in the string shown in the figure is ........ $N$.

$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module