A block of mass 4 kg is at rest on a rough i | vedclass

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

Question

(B) Given: Mass $m = 4 \ kg$, coefficient of static friction $\mu_s = 0.5$, and frictional force $f = 14.14 \ N = 10\sqrt{2} \ N$.Since the block is a

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given: Mass $m = 4 \ kg$, coefficient of static friction $\mu_s = 0.5$, and frictional force $f = 14.14 \ N = 10\sqrt{2} \ N$.Since the block is at rest on the inclined plane, the frictional force balances the component of gravitational force acting down the plane.$f = mg \sin 	heta$$10\sqrt{2} = 4 	imes 10 	imes \sin 	heta$$10\sqrt{2} = 40 \sin 	heta$$\sin 	heta = \frac{10\sqrt{2}}{40} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}} \approx 0.3535$.However, if we assume the frictional force given is the limiting friction $(f_L = \mu_s N)$:Normal reaction $N = mg \cos 	heta = 40 \cos 	heta$.$f_L = \mu_s N = 0.5 	imes 40 \cos 	heta = 20 \cos 	heta$.Equating $f_L = 10\sqrt{2}$:$20 \cos 	heta = 10\sqrt{2}$$\cos 	heta = \frac{10\sqrt{2}}{20} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore, $	heta = 45^{\circ}$.

$A$ block of mass $4 \ kg$ is at rest on a rough inclined plane making an angle of $\theta$ with the horizontal. The coefficient of static friction between the block and the plane is $0.5$ and the frictional force on the block is $14.14 \ N$. Find the value of $\theta$. (in $^{\circ}$)

Similar Questions

$A$ block is kept on an inclined plane of inclination $\theta$ and length $l$. The velocity of the block at the bottom of the inclined plane is (the coefficient of friction is $\mu$):

$A$ block is projected up an inclined plane of angle $\theta = 30^o$ with an initial velocity of $5 \, m/s$. It comes to rest in $0.5 \, s$. What is the coefficient of friction?

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

Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

Vedclass Test Series

Exam Paper Generator

Online Exam Module