Two blocks A and B of equal masses are slidin | vedclass

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

Question

(A) The acceleration of a block sliding down an inclined plane is given by $a = g(\sin 	heta - \mu \cos 	heta)$.For block $A$: $a_A = 10(\sin 45^{\c

Vedclass · Accepted Answer

$2 	ext{ s}, 8\sqrt{2} 	ext{ m}$

Vedclass · Answer

(A) The acceleration of a block sliding down an inclined plane is given by $a = g(\sin 	heta - \mu \cos 	heta)$.For block $A$: $a_A = 10(\sin 45^{\circ} - 0.2 \cos 45^{\circ}) = 10 	imes \frac{1}{\sqrt{2}}(1 - 0.2) = \frac{8}{\sqrt{2}} 	ext{ m/s}^2$.For block $B$: $a_B = 10(\sin 45^{\circ} - 0.3 \cos 45^{\circ}) = 10 	imes \frac{1}{\sqrt{2}}(1 - 0.3) = \frac{7}{\sqrt{2}} 	ext{ m/s}^2$.The relative acceleration of $A$ with respect to $B$ is $a_{AB} = a_A - a_B = \frac{8}{\sqrt{2}} - \frac{7}{\sqrt{2}} = \frac{1}{\sqrt{2}} 	ext{ m/s}^2$.Given the initial separation $s_{AB} = \sqrt{2} 	ext{ m}$,we use the equation of motion $s_{AB} = u_{AB}t + \frac{1}{2}a_{AB}t^2$. Since $u_{AB} = 0$,we have $\sqrt{2} = \frac{1}{2} 	imes \frac{1}{\sqrt{2}} 	imes t^2$.Solving for $t$: $t^2 = 4 \Rightarrow t = 2 	ext{ s}$.Now,the distance traveled by block $A$ from its initial position is $s_A = \frac{1}{2}a_A t^2 = \frac{1}{2} 	imes \frac{8}{\sqrt{2}} 	imes (2)^2 = \frac{1}{2} 	imes \frac{8}{\sqrt{2}} 	imes 4 = \frac{16}{\sqrt{2}} = 8\sqrt{2} 	ext{ m}$.

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 $\mu$):

$A$ body of mass $m$ is launched up on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$. If the time of ascent is half of the time of descent,the value of $x$ is ..... .

$A$ block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

If $250 \,J$ of work is done in sliding a $5 \,kg$ block up an inclined plane of height $4 \,m$,the value of work done against friction is ............. $J$ $\left(g=10 \,ms^{-2}\right)$.

$A$ body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is $0.5$. Then its retardation is ($g$ = acceleration due to gravity).

Vedclass Test Series

Exam Paper Generator

Online Exam Module