A block of mass 15, kg is resting on a rough | vedclass

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

Question

(A) The block is in equilibrium. Let the angle of inclination be $	heta = 45^{\circ}$. The mass of the block is $m = 15\, kg$,so the weight is $W = m

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) The block is in equilibrium. Let the angle of inclination be $	heta = 45^{\circ}$. The mass of the block is $m = 15\, kg$,so the weight is $W = mg = 15 	imes 10 = 150\, N$.We resolve the forces along the inclined plane ($x$-axis) and perpendicular to it ($y$-axis).The forces acting on the block are:$1$. Weight $(mg = 150\, N)$ acting vertically downwards.$2$. Normal reaction $(N)$ perpendicular to the plane.$3$. Tension $(T = 50\, N)$ acting horizontally.$4$. Frictional force $(f = \mu N)$ acting down the plane to oppose the tendency of motion.Resolving forces along the $x$-axis (parallel to the plane):$\Sigma F_x = 0 \implies T \cos 45^{\circ} = mg \sin 45^{\circ} + f$$50 \cos 45^{\circ} = 150 \sin 45^{\circ} - f$$f = 150 \sin 45^{\circ} - 50 \cos 45^{\circ} = (150 - 50) 	imes \frac{1}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, N$.Resolving forces along the $y$-axis (perpendicular to the plane):$\Sigma F_y = 0 \implies N = mg \cos 45^{\circ} + T \sin 45^{\circ}$$N = 150 \cos 45^{\circ} + 50 \sin 45^{\circ} = (150 + 50) 	imes \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2}\, N$.Using $f = \mu N$:$\mu = \frac{f}{N} = \frac{50\sqrt{2}}{100\sqrt{2}} = \frac{50}{100} = \frac{1}{2}$.

$A$ block of mass $15\, kg$ is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal string which has a tension of $50\, N$. The coefficient of friction between the surfaces of contact is $(g = 10\, m/s^2)$.

Similar Questions

$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

$A$ 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:

An object takes $n$ times as much time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down a perfectly smooth inclined plane of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by:

As shown in the figure,a body of mass $m$ moving vertically downward with a speed of $3\, m/s$ hits a smooth fixed inclined plane and rebounds with a velocity $v_f$ in the horizontal direction. If the angle of the inclined plane is $30^{\circ}$,the velocity $v_f$ will be:

$A$ brick of mass $2 \, kg$ begins to slide down on a plane inclined at an angle of $45^\circ$ with the horizontal. The force of friction will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module