In the above question,the speed of the mass w | vedclass

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

Question

(A) The acceleration of the mass is given by $a = g(\sin 	heta - \mu_0 x \cos 	heta)$.Using the work-energy theorem or integrating $v dv = a dx$,we

Vedclass · Accepted Answer

$\sqrt{\frac{g 	an 	heta \sin 	heta}{\mu_0}}$

Vedclass · Answer

(A) The acceleration of the mass is given by $a = g(\sin 	heta - \mu_0 x \cos 	heta)$.Using the work-energy theorem or integrating $v dv = a dx$,we have $\int_0^{v} v dv = \int_0^{x/2} g(\sin 	heta - \mu_0 x \cos 	heta) dx$.Integrating both sides: $\frac{v^2}{2} = g[\sin 	heta \cdot x' - \frac{\mu_0}{2} (x')^2 \cos 	heta]$,where $x' = x/2$.Substituting $x' = \frac{x}{2} = \frac{1}{\mu_0} 	an 	heta$ (since maximum distance $x_{max} = \frac{2 	an 	heta}{\mu_0}$):$\frac{v^2}{2} = g[\sin 	heta (\frac{	an 	heta}{\mu_0}) - \frac{\mu_0}{2} (\frac{	an 	heta}{\mu_0})^2 \cos 	heta]$.$\frac{v^2}{2} = g[\frac{\sin 	heta 	an 	heta}{\mu_0} - \frac{	an^2 	heta \cos 	heta}{2 \mu_0}] = g[\frac{\sin 	heta 	an 	heta}{\mu_0} - \frac{\sin 	heta 	an 	heta}{2 \mu_0}] = \frac{g \sin 	heta 	an 	heta}{2 \mu_0}$.Thus,$v^2 = \frac{g \sin 	heta 	an 	heta}{\mu_0}$,which gives $v = \sqrt{\frac{g 	an 	heta \sin 	heta}{\mu_0}}$.

In the above question,the speed of the mass when it has traveled half the maximum distance is:

Similar Questions

$A$ 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:

$A$ 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 minimum coefficient of friction between the surfaces of contact is $(g = 10 \, m/s^2)$.

$A$ block of mass $m$ slides down a plane inclined at an angle $\theta$. Which of the following will $NOT$ increase the energy lost by the block due to friction?

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

$A$ block of mass $10\, kg$ is placed on an inclined plane. When the angle of inclination is $30^\circ$,the block just begins to slide down the plane. The force of static friction is ....... $kg\, wt$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module