An inclined plane of length 5.60 m making an | vedclass

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

Question

(B) Given: $	heta = 45^{\circ}$,$d = 5.60 \ m$,$E = 100 \ V \ m^{-1}$,$m = 1 \ kg$,$\mu = 0.1$,$q = 10^{-2} \ C$,$v_0 = 0$.From the free body diagram

Vedclass · Accepted Answer

$1.41 \ s$

Vedclass · Answer

(B) Given: $	heta = 45^{\circ}$,$d = 5.60 \ m$,$E = 100 \ V \ m^{-1}$,$m = 1 \ kg$,$\mu = 0.1$,$q = 10^{-2} \ C$,$v_0 = 0$.From the free body diagram,the normal force $N$ is:$N = mg \cos 45^{\circ} + qE \sin 45^{\circ}$$N = (1 	imes 10 	imes \frac{1}{\sqrt{2}}) + (10^{-2} 	imes 100 	imes \frac{1}{\sqrt{2}})$$N = \frac{10}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ N$The net force $F$ acting down the incline is:$F = mg \sin 45^{\circ} - qE \cos 45^{\circ} - \mu N$$F = (1 	imes 10 	imes \frac{1}{\sqrt{2}}) - (10^{-2} 	imes 100 	imes \frac{1}{\sqrt{2}}) - (0.1 	imes 7.778)$$F = \frac{10}{\sqrt{2}} - \frac{1}{\sqrt{2}} - 0.7778 = \frac{9}{\sqrt{2}} - 0.7778 \approx 6.364 - 0.778 = 5.586 \ N$Acceleration $a = \frac{F}{m} = \frac{5.586}{1} = 5.586 \ m \ s^{-2}$.Using $d = v_0 t + \frac{1}{2} a t^2$:$5.60 = 0 + \frac{1}{2} 	imes 5.586 	imes t^2$$t^2 = \frac{2 	imes 5.60}{5.586} \approx 2$$t = \sqrt{2} \approx 1.41 \ s$.

Similar Questions

$A$ block kept on a rough surface starts sliding when the inclination of the surface is $\theta$ with respect to the horizontal. The coefficient of static friction between the block and the surface is

$A$ cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$. The value of the coefficient of kinetic friction is:

$A$ block of mass $2\,\,kg$ is placed on a rough inclined plane as shown in the figure $(\mu = 0.2)$ so that it just touches the spring. The block is allowed to move downwards. The spring will be compressed to a maximum of

$A$ body of mass $2\,kg$ slides down with an acceleration of $3\,m/s^2$ on a rough inclined plane having a slope of $30^o$. The external force required to take the same body up the plane with the same acceleration will be ........ $N$ $(g = 10\,m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module