A block rests on a rough inclined plane makin | vedclass

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

Question

(A) The angle of repose is given by $\alpha = 	an^{-1}(\mu) = 	an^{-1}(0.8) \approx 38.6^{\circ}$.Since the angle of the inclined plane $	heta = 30

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The angle of repose is given by $\alpha = 	an^{-1}(\mu) = 	an^{-1}(0.8) \approx 38.6^{\circ}$.Since the angle of the inclined plane $	heta = 30^{\circ}$ is less than the angle of repose $\alpha$,the block remains at rest.For a block at rest on an inclined plane,the static frictional force $f_s$ balances the component of the gravitational force acting down the plane.Thus,$f_s = mg \sin(	heta)$.Given $f_s = 10 \, N$,$g = 10 \, m/s^2$,and $	heta = 30^{\circ}$,we have:$10 = m 	imes 10 	imes \sin(30^{\circ})$$10 = m 	imes 10 	imes 0.5$$10 = 5m$$m = \frac{10}{5} = 2 \, kg$.

$A$ 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 $g = 10 \, m/s^2$)

Similar Questions

$A$ wire carrying a current $i$ is placed in a uniform magnetic field in the form of the curve $y = a \sin \left( \frac{\pi x}{L} \right)$ for $0 \le x \le 2L$. The force acting on the wire is

Which of the following is incorrect regarding $H_2O_2$?

Out of molar entropy $(I)$,specific volume $(II)$,heat capacity $(III)$,volume $(IV)$,which are extensive properties?

$A$ transformer works on the principle of

Diazo-coupling is useful to prepare some

Vedclass Test Series

Exam Paper Generator

Online Exam Module