Two blocks of masses 1 kg and 2 kg are conn | vedclass

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

Question

(B) The frictional force acting on both blocks is kinetic in nature.For the $1 \ kg$ block:$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 	imes 1 	imes 10 \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The frictional force acting on both blocks is kinetic in nature.For the $1 \ kg$ block:$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 	imes 1 	imes 10 	imes \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ N$For the $2 \ kg$ block:$f_2 = \mu m_2 g \cos 45^{\circ} = 0.4 	imes 2 	imes 10 	imes \frac{1}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \ N$The net force along the incline in the downward direction provides the acceleration $a = \alpha \sqrt{2}$:$(m_1 + m_2)g \sin 45^{\circ} - (f_1 + f_2) = (m_1 + m_2)a$$(1 + 2) 	imes 10 	imes \frac{1}{\sqrt{2}} - (2\sqrt{2} + 4\sqrt{2}) = (1 + 2) 	imes (\alpha \sqrt{2})$$30 	imes \frac{1}{\sqrt{2}} - 6\sqrt{2} = 3\alpha \sqrt{2}$Multiply throughout by $\sqrt{2}$:$30 - 6 	imes 2 = 3\alpha 	imes 2$$30 - 12 = 6\alpha$$18 = 6\alpha$$\alpha = 3$

Similar Questions

$A$ particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu = \frac{1}{\sqrt{3}}$. The maximum height up to which the particle can remain stationary is

$Assertion$ : Angle of repose is equal to the angle of limiting friction.
$Reason$ : When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

$A$ uniform chain of length $L$ which hangs partially from a table,is kept in equilibrium by friction. The maximum length that can hang without slipping is $l$. Then,the coefficient of friction between the table and the chain is:

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$,then the angle of the inclined plane is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module