A block of mass m is placed on a surface with | vedclass

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

Question

(D) The condition for the block not to slip on an inclined surface is that the angle of inclination $	heta$ must be less than or equal to the angle o

Vedclass · Accepted Answer

$\frac{1}{6} \ m$

Vedclass · Answer

(D) The condition for the block not to slip on an inclined surface is that the angle of inclination $	heta$ must be less than or equal to the angle of repose $\phi$,where $	an \phi = \mu$.Thus,for the limiting case,$	an 	heta = \mu$.The slope of the surface at any point $x$ is given by $\frac{dy}{dx} = 	an 	heta$.Given $y = \frac{x^3}{6}$,we have $\frac{dy}{dx} = \frac{3x^2}{6} = \frac{x^2}{2}$.Equating the slope to the coefficient of friction $\mu = 0.5$:$\frac{x^2}{2} = 0.5$$x^2 = 1$$x = 1$ (considering the positive side).Now,substitute $x = 1$ into the equation of the surface to find the maximum height $y$:$y = \frac{x^3}{6} = \frac{1^3}{6} = \frac{1}{6} \ m$.

$A$ block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

Similar Questions

$A$ person climbs up a conveyor belt with a constant acceleration. The speed of the belt is $\sqrt{\frac{g h}{6}}$ and the coefficient of friction is $\frac{5}{3 \sqrt{3}}$. The time taken by the person to reach from $A$ to $B$ with the maximum possible acceleration is

$A$ small mass slides down an inclined plane of angle $\theta$ with the horizontal. The coefficient of friction is $\mu = \mu_0 x$,where $x$ is the distance through which the mass slides down and $\mu_0$ is a constant. Then,the distance covered by the mass before it stops is:

If $250 \,J$ of work is done in sliding a $5 \,kg$ block up an inclined plane of height $4 \,m$,the value of work done against friction is ............. $J$ $\left(g=10 \,ms^{-2}\right)$.

$A$ block of mass $1 \,kg$ slides down on a rough inclined plane of inclination $60^\circ$ starting from its top. If the coefficient of kinetic friction is $0.5$ and the length of the plane is $1 \,m$,then the work done against friction is ........ $J$.

$A$ body of $5 \,kg$ weight kept on a rough inclined plane of angle $30^{\circ}$ starts sliding with a constant velocity. Then the coefficient of friction is (assume $g=10 \,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module