Consider a small block sliding down an inclin | vedclass

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

Question

(B) The normal force on the block is $N = mg \cos 	heta$.Since the coefficient of friction is $\mu = \frac{2}{3} x$, the kinetic friction force is $f

Vedclass · Accepted Answer

$\sqrt{3} 	ext{ m}$

Vedclass · Answer

(B) The normal force on the block is $N = mg \cos 	heta$.Since the coefficient of friction is $\mu = \frac{2}{3} x$, the kinetic friction force is $f_k = \mu N = \left( \frac{2}{3} x ight) mg \cos 	heta$.According to the work-energy theorem, the total work done on the block is equal to the change in its kinetic energy: $W_g + W_f = \Delta K.E. = 0 - 0 = 0$.The work done by gravity as the block slides down a distance $x$ is $W_g = mgx \sin 	heta$.The work done by the variable friction force is $W_f = - \int_0^x f_k \, dx = - \int_0^x \left( \frac{2}{3} x mg \cos 	heta ight) dx = - \frac{2}{3} mg \cos 	heta \left[ \frac{x^2}{2} ight]_0^x = - \frac{1}{3} mgx^2 \cos 	heta$.Substituting these into the work-energy equation: $mgx \sin 	heta - \frac{1}{3} mgx^2 \cos 	heta = 0$.Dividing by $mgx$ (assuming $x 
eq 0$): $\sin 	heta - \frac{1}{3} x \cos 	heta = 0$.Solving for $x$: $x = 3 	an 	heta$.Given $	heta = 30^{\circ}$, $x = 3 	an 30^{\circ} = 3 	imes \frac{1}{\sqrt{3}} = \sqrt{3} 	ext{ m}$.

Consider a small block sliding down an inclined plane of inclination $30^{\circ}$ with the horizontal. The coefficient of friction is $\mu = \frac{2}{3} x$, where $x$ is the distance (in meters) through which the mass slides down. The distance covered by the mass before it stops is

Similar Questions

The ratio of acceleration of block $A$ placed on a smooth incline to block $B$ placed on a rough incline is $2:1$. The coefficient of kinetic friction between block $B$ and the incline is .........

What percentage of the total length of a chain can hang off the edge of a table without sliding,given that the coefficient of friction between the chain and the table is $0.25$?

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

$A$ car is going at a speed of $6\, m/s$ when it encounters a $15\, m$ slope of angle $30^o$. The friction coefficient between the road and tyre is $0.5$. The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ........ $m/s$. $(g = 10\, m/s^2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module