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Laws of Motion

146158 Two touching blocks 1 and 2 are placed on an inclined plane forming an angle \(60^{\circ}\) with the horizontal. The masses are \(m_{1}\) and \(m_{2}\) and the coefficient of friction between the inclined plane and the two blocks are \(1.5 \mu\) and \(1.0 \mu\), respectively. The force of reaction between the blocks during the motion is \((\mathrm{g}=\) acceleration due to gravity)

1 \(\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mu \mathrm{g}\)

2 \(\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mu \mathrm{g}\)

3 \(\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

4 \(\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

Explanation:

D Given,
Coefficient of friction for first block \(\left(\mu_{1}\right)=1.5 \mu\)
Coefficient of friction for second block \(\left(\mu_{2}\right)=1 \mu\)
Acceleration of both block \(=\mathrm{a}\)

Now, FBD for the first block,

If a be the common acceleration, then
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-\mu_{1} \cdot \mathrm{R}_{1}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-1.5 \mu \cdot \mathrm{m}_{1} \mathrm{~g} \cos 60^{\circ}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \frac{\sqrt{3}}{2}-1.5 \mu \mathrm{m}_{1} \mathrm{~g} \times \frac{1}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
Similarly, for second block,

Since, \(\quad \mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu_{2} \mathrm{R}_{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu \mathrm{m}_{2} \mathrm{~g} \cos 60^{\circ}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\mu \mathrm{m}_{2} \mathrm{~g} \times \frac{1}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
Dividing equation (i) by (ii), we get -
\(\frac{\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}}{\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}}=\frac{\mathrm{m}_{1} \mathrm{a}}{\mathrm{m}_{2} \mathrm{a}}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~g}+2 \mathrm{R}}{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{2} \mathrm{~g}-2 \mathrm{R}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\)
\(\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}+2 \mathrm{Rm}_{2}\)
\(=\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}-2 \mathrm{Rm}_{1}\)
\(2 \mathrm{R}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=0.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}\)
\(\mathrm{R}=\frac{\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}}{4\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\)
\(\mathrm{R}=\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

TS- EAMCET-03.05.2019

Laws of Motion

146159 Consider a system of blocks \(X, A\) and \(B\) as shown in the figure. The blocks \(A\) and \(B\) have equal mass and are connected by a mass-less string through a mass-less pulley. The coefficient of friction between block \(A\) and \(X\) or \(B\) and \(X\) is 0.5. If block \(X\) moves on the horizontal frictionless surface what should be its minimum acceleration such that blocks \(A\) and \(B\) remain stationary, \((g=\) acceleration due to gravity.)

1 \(\frac{g}{3}\)

2 \(3 \mathrm{~g}\)

3 \(\frac{g}{4}\)

4 \(\frac{3 g}{4}\)

TS- EAMCET-03.05.2019

Laws of Motion

146160 A block of mass \(10 \mathrm{~kg}\), initially at rest, makes a downward motion on \(45^{\circ}\) inclined plane. Then the distance travelled by the block after \(2 \mathrm{~s}\) is (Assume the coefficient of kinetic friction to be 0.3 and \(g=10 \mathrm{~ms}^{-2}\) )

1 \(7 \sqrt{2} \mathrm{~m}\)

2 \(\frac{9}{\sqrt{2}} \mathrm{~m}\)

3 \(10 \sqrt{2} \mathrm{~m}\)

4 \(5 \sqrt{2} \mathrm{~m}\)

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane \(\mathrm{BCE}\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular wooden block ABCD of height \(10 \mathrm{~m}\), as shown in the figure. \(A\) small block is allowed to slide down from the top \(E\) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{8}\) and the angle of inclination of the inclined plane is \(\sin ^{-1}(0.6)\). If the small block finally reaches the ground at a point \(F\), then DF will be
(Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{5}{3} \mathrm{~m}\)

2 \(\frac{10}{3} \mathrm{~m}\)

3 \(\frac{13}{3} \mathrm{~m}\)

4 \(\frac{20}{3} \mathrm{~m}\)

Explanation:

D Angle of inclination \(\theta=\sin ^{-1}(0.6)\) \(\sin \theta=0.6\)
\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-(0.6)^{2}}\)
\(\cos \theta=0.8\)

Given, \(\mu_{\mathrm{k}}=\frac{1}{8}\)
Mass of block \(=m\)
Let, acceleration of mass \(=\mathrm{a}\)
Block slide down therefore
Net force on block (downward)
\(\mathrm{R}=\mathrm{mg} \cos \theta\)
\(\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma} \quad\) (downward)
\(\mathrm{mg} \sin \theta-\mu_{\mathrm{k}} \mathrm{mg} \cos \theta=\mathrm{ma}\)
\((\therefore \mathrm{f}=\mu \mathrm{R})\)
\(\mathrm{g} \sin \theta-\mu_{\mathrm{k}} \mathrm{g} \cos \theta=\mathrm{a}\)
\(10 \times 0.6-\frac{1}{8} \times 10 \times 0.8=a\)
\(6-1=a\)
\(a=5 \mathrm{~m} / \mathrm{sec}^{2}\)
Velocity of mass at point \(\mathrm{C}\)
\(v^{2} =u^{2}+2 a s\)
\(v^{2} =0+2 \times 5 \times s\)
\(\mathrm{v}^{2}=10 \mathrm{~s}(\therefore \mathrm{u}=0, \text { velocity at } \mathrm{E}=0)\)
Here, \(\mathrm{s}=\mathrm{EC}\)
In triangle \(\mathrm{ECB}\)
\(\sin \theta=\frac{\mathrm{BE}}{\mathrm{EC}},\left(\mathrm{BE}=\frac{25}{6}\right)\)
\(0.6=\frac{\frac{25}{6}}{\mathrm{EC}}\)
\(\mathrm{EC} =\frac{25}{3.6}\)
\(\therefore \quad \mathrm{v}^{2} =10 \mathrm{~s}\)
\(\mathrm{v}^{2} =10 \times \frac{25}{3.6}=10 \times \frac{25}{36} \times 10\)
\(\mathrm{v} =\frac{10 \times 5}{6}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
\(\mathrm{v} =\frac{25}{3} \mathrm{~m} / \mathrm{sec}\)
At point \(\mathrm{C}\) we draw velocity component. In vertical direction-
\(\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\)
here
\(\mathrm{h}=10\)
\(\mathrm{u}=\mathrm{v} \sin \theta=\frac{25}{3} \times 0.6\)
Then,
\(10=\frac{25}{3} \times 0.6 \mathrm{t}+\frac{1}{2} \times 10 \times \mathrm{t}^{2}\)
\(10=5 \mathrm{t}+5 \mathrm{t}^{2}\)
\(\mathrm{t}^{2}+\mathrm{t}-2=0\)
\((\mathrm{t}+2)(\mathrm{t}-1)=0\)
\(\mathrm{t}=1 \sec\)
\(\mathrm{DF}=(\) Horizontal component of velocity \() \times\) time
\(\mathrm{DF}=\mathrm{v} \cos \theta \times \mathrm{t}\)
\(=\frac{25}{3} \times 0.8 \times 1\)
\(\mathrm{DF}=\frac{20}{3} \mathrm{~m}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146158 Two touching blocks 1 and 2 are placed on an inclined plane forming an angle \(60^{\circ}\) with the horizontal. The masses are \(m_{1}\) and \(m_{2}\) and the coefficient of friction between the inclined plane and the two blocks are \(1.5 \mu\) and \(1.0 \mu\), respectively. The force of reaction between the blocks during the motion is \((\mathrm{g}=\) acceleration due to gravity)

1 \(\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mu \mathrm{g}\)

2 \(\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mu \mathrm{g}\)

3 \(\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

4 \(\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

Explanation:

D Given,
Coefficient of friction for first block \(\left(\mu_{1}\right)=1.5 \mu\)
Coefficient of friction for second block \(\left(\mu_{2}\right)=1 \mu\)
Acceleration of both block \(=\mathrm{a}\)

Now, FBD for the first block,

If a be the common acceleration, then
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-\mu_{1} \cdot \mathrm{R}_{1}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-1.5 \mu \cdot \mathrm{m}_{1} \mathrm{~g} \cos 60^{\circ}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \frac{\sqrt{3}}{2}-1.5 \mu \mathrm{m}_{1} \mathrm{~g} \times \frac{1}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
Similarly, for second block,

Since, \(\quad \mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu_{2} \mathrm{R}_{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu \mathrm{m}_{2} \mathrm{~g} \cos 60^{\circ}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\mu \mathrm{m}_{2} \mathrm{~g} \times \frac{1}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
Dividing equation (i) by (ii), we get -
\(\frac{\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}}{\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}}=\frac{\mathrm{m}_{1} \mathrm{a}}{\mathrm{m}_{2} \mathrm{a}}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~g}+2 \mathrm{R}}{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{2} \mathrm{~g}-2 \mathrm{R}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\)
\(\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}+2 \mathrm{Rm}_{2}\)
\(=\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}-2 \mathrm{Rm}_{1}\)
\(2 \mathrm{R}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=0.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}\)
\(\mathrm{R}=\frac{\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}}{4\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\)
\(\mathrm{R}=\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

TS- EAMCET-03.05.2019

Laws of Motion

146159 Consider a system of blocks \(X, A\) and \(B\) as shown in the figure. The blocks \(A\) and \(B\) have equal mass and are connected by a mass-less string through a mass-less pulley. The coefficient of friction between block \(A\) and \(X\) or \(B\) and \(X\) is 0.5. If block \(X\) moves on the horizontal frictionless surface what should be its minimum acceleration such that blocks \(A\) and \(B\) remain stationary, \((g=\) acceleration due to gravity.)

1 \(\frac{g}{3}\)

2 \(3 \mathrm{~g}\)

3 \(\frac{g}{4}\)

4 \(\frac{3 g}{4}\)

TS- EAMCET-03.05.2019

Laws of Motion

146160 A block of mass \(10 \mathrm{~kg}\), initially at rest, makes a downward motion on \(45^{\circ}\) inclined plane. Then the distance travelled by the block after \(2 \mathrm{~s}\) is (Assume the coefficient of kinetic friction to be 0.3 and \(g=10 \mathrm{~ms}^{-2}\) )

1 \(7 \sqrt{2} \mathrm{~m}\)

2 \(\frac{9}{\sqrt{2}} \mathrm{~m}\)

3 \(10 \sqrt{2} \mathrm{~m}\)

4 \(5 \sqrt{2} \mathrm{~m}\)

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane \(\mathrm{BCE}\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular wooden block ABCD of height \(10 \mathrm{~m}\), as shown in the figure. \(A\) small block is allowed to slide down from the top \(E\) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{8}\) and the angle of inclination of the inclined plane is \(\sin ^{-1}(0.6)\). If the small block finally reaches the ground at a point \(F\), then DF will be
(Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{5}{3} \mathrm{~m}\)

2 \(\frac{10}{3} \mathrm{~m}\)

3 \(\frac{13}{3} \mathrm{~m}\)

4 \(\frac{20}{3} \mathrm{~m}\)

Explanation:

D Angle of inclination \(\theta=\sin ^{-1}(0.6)\) \(\sin \theta=0.6\)
\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-(0.6)^{2}}\)
\(\cos \theta=0.8\)

Given, \(\mu_{\mathrm{k}}=\frac{1}{8}\)
Mass of block \(=m\)
Let, acceleration of mass \(=\mathrm{a}\)
Block slide down therefore
Net force on block (downward)
\(\mathrm{R}=\mathrm{mg} \cos \theta\)
\(\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma} \quad\) (downward)
\(\mathrm{mg} \sin \theta-\mu_{\mathrm{k}} \mathrm{mg} \cos \theta=\mathrm{ma}\)
\((\therefore \mathrm{f}=\mu \mathrm{R})\)
\(\mathrm{g} \sin \theta-\mu_{\mathrm{k}} \mathrm{g} \cos \theta=\mathrm{a}\)
\(10 \times 0.6-\frac{1}{8} \times 10 \times 0.8=a\)
\(6-1=a\)
\(a=5 \mathrm{~m} / \mathrm{sec}^{2}\)
Velocity of mass at point \(\mathrm{C}\)
\(v^{2} =u^{2}+2 a s\)
\(v^{2} =0+2 \times 5 \times s\)
\(\mathrm{v}^{2}=10 \mathrm{~s}(\therefore \mathrm{u}=0, \text { velocity at } \mathrm{E}=0)\)
Here, \(\mathrm{s}=\mathrm{EC}\)
In triangle \(\mathrm{ECB}\)
\(\sin \theta=\frac{\mathrm{BE}}{\mathrm{EC}},\left(\mathrm{BE}=\frac{25}{6}\right)\)
\(0.6=\frac{\frac{25}{6}}{\mathrm{EC}}\)
\(\mathrm{EC} =\frac{25}{3.6}\)
\(\therefore \quad \mathrm{v}^{2} =10 \mathrm{~s}\)
\(\mathrm{v}^{2} =10 \times \frac{25}{3.6}=10 \times \frac{25}{36} \times 10\)
\(\mathrm{v} =\frac{10 \times 5}{6}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
\(\mathrm{v} =\frac{25}{3} \mathrm{~m} / \mathrm{sec}\)
At point \(\mathrm{C}\) we draw velocity component. In vertical direction-
\(\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\)
here
\(\mathrm{h}=10\)
\(\mathrm{u}=\mathrm{v} \sin \theta=\frac{25}{3} \times 0.6\)
Then,
\(10=\frac{25}{3} \times 0.6 \mathrm{t}+\frac{1}{2} \times 10 \times \mathrm{t}^{2}\)
\(10=5 \mathrm{t}+5 \mathrm{t}^{2}\)
\(\mathrm{t}^{2}+\mathrm{t}-2=0\)
\((\mathrm{t}+2)(\mathrm{t}-1)=0\)
\(\mathrm{t}=1 \sec\)
\(\mathrm{DF}=(\) Horizontal component of velocity \() \times\) time
\(\mathrm{DF}=\mathrm{v} \cos \theta \times \mathrm{t}\)
\(=\frac{25}{3} \times 0.8 \times 1\)
\(\mathrm{DF}=\frac{20}{3} \mathrm{~m}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146158 Two touching blocks 1 and 2 are placed on an inclined plane forming an angle \(60^{\circ}\) with the horizontal. The masses are \(m_{1}\) and \(m_{2}\) and the coefficient of friction between the inclined plane and the two blocks are \(1.5 \mu\) and \(1.0 \mu\), respectively. The force of reaction between the blocks during the motion is \((\mathrm{g}=\) acceleration due to gravity)

1 \(\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mu \mathrm{g}\)

2 \(\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mu \mathrm{g}\)

3 \(\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

4 \(\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

Explanation:

D Given,
Coefficient of friction for first block \(\left(\mu_{1}\right)=1.5 \mu\)
Coefficient of friction for second block \(\left(\mu_{2}\right)=1 \mu\)
Acceleration of both block \(=\mathrm{a}\)

Now, FBD for the first block,

If a be the common acceleration, then
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-\mu_{1} \cdot \mathrm{R}_{1}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-1.5 \mu \cdot \mathrm{m}_{1} \mathrm{~g} \cos 60^{\circ}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \frac{\sqrt{3}}{2}-1.5 \mu \mathrm{m}_{1} \mathrm{~g} \times \frac{1}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
Similarly, for second block,

Since, \(\quad \mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu_{2} \mathrm{R}_{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu \mathrm{m}_{2} \mathrm{~g} \cos 60^{\circ}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\mu \mathrm{m}_{2} \mathrm{~g} \times \frac{1}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
Dividing equation (i) by (ii), we get -
\(\frac{\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}}{\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}}=\frac{\mathrm{m}_{1} \mathrm{a}}{\mathrm{m}_{2} \mathrm{a}}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~g}+2 \mathrm{R}}{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{2} \mathrm{~g}-2 \mathrm{R}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\)
\(\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}+2 \mathrm{Rm}_{2}\)
\(=\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}-2 \mathrm{Rm}_{1}\)
\(2 \mathrm{R}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=0.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}\)
\(\mathrm{R}=\frac{\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}}{4\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\)
\(\mathrm{R}=\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

TS- EAMCET-03.05.2019

Laws of Motion

146159 Consider a system of blocks \(X, A\) and \(B\) as shown in the figure. The blocks \(A\) and \(B\) have equal mass and are connected by a mass-less string through a mass-less pulley. The coefficient of friction between block \(A\) and \(X\) or \(B\) and \(X\) is 0.5. If block \(X\) moves on the horizontal frictionless surface what should be its minimum acceleration such that blocks \(A\) and \(B\) remain stationary, \((g=\) acceleration due to gravity.)

1 \(\frac{g}{3}\)

2 \(3 \mathrm{~g}\)

3 \(\frac{g}{4}\)

4 \(\frac{3 g}{4}\)

TS- EAMCET-03.05.2019

Laws of Motion

146160 A block of mass \(10 \mathrm{~kg}\), initially at rest, makes a downward motion on \(45^{\circ}\) inclined plane. Then the distance travelled by the block after \(2 \mathrm{~s}\) is (Assume the coefficient of kinetic friction to be 0.3 and \(g=10 \mathrm{~ms}^{-2}\) )

1 \(7 \sqrt{2} \mathrm{~m}\)

2 \(\frac{9}{\sqrt{2}} \mathrm{~m}\)

3 \(10 \sqrt{2} \mathrm{~m}\)

4 \(5 \sqrt{2} \mathrm{~m}\)

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane \(\mathrm{BCE}\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular wooden block ABCD of height \(10 \mathrm{~m}\), as shown in the figure. \(A\) small block is allowed to slide down from the top \(E\) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{8}\) and the angle of inclination of the inclined plane is \(\sin ^{-1}(0.6)\). If the small block finally reaches the ground at a point \(F\), then DF will be
(Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{5}{3} \mathrm{~m}\)

2 \(\frac{10}{3} \mathrm{~m}\)

3 \(\frac{13}{3} \mathrm{~m}\)

4 \(\frac{20}{3} \mathrm{~m}\)

Explanation:

D Angle of inclination \(\theta=\sin ^{-1}(0.6)\) \(\sin \theta=0.6\)
\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-(0.6)^{2}}\)
\(\cos \theta=0.8\)

Given, \(\mu_{\mathrm{k}}=\frac{1}{8}\)
Mass of block \(=m\)
Let, acceleration of mass \(=\mathrm{a}\)
Block slide down therefore
Net force on block (downward)
\(\mathrm{R}=\mathrm{mg} \cos \theta\)
\(\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma} \quad\) (downward)
\(\mathrm{mg} \sin \theta-\mu_{\mathrm{k}} \mathrm{mg} \cos \theta=\mathrm{ma}\)
\((\therefore \mathrm{f}=\mu \mathrm{R})\)
\(\mathrm{g} \sin \theta-\mu_{\mathrm{k}} \mathrm{g} \cos \theta=\mathrm{a}\)
\(10 \times 0.6-\frac{1}{8} \times 10 \times 0.8=a\)
\(6-1=a\)
\(a=5 \mathrm{~m} / \mathrm{sec}^{2}\)
Velocity of mass at point \(\mathrm{C}\)
\(v^{2} =u^{2}+2 a s\)
\(v^{2} =0+2 \times 5 \times s\)
\(\mathrm{v}^{2}=10 \mathrm{~s}(\therefore \mathrm{u}=0, \text { velocity at } \mathrm{E}=0)\)
Here, \(\mathrm{s}=\mathrm{EC}\)
In triangle \(\mathrm{ECB}\)
\(\sin \theta=\frac{\mathrm{BE}}{\mathrm{EC}},\left(\mathrm{BE}=\frac{25}{6}\right)\)
\(0.6=\frac{\frac{25}{6}}{\mathrm{EC}}\)
\(\mathrm{EC} =\frac{25}{3.6}\)
\(\therefore \quad \mathrm{v}^{2} =10 \mathrm{~s}\)
\(\mathrm{v}^{2} =10 \times \frac{25}{3.6}=10 \times \frac{25}{36} \times 10\)
\(\mathrm{v} =\frac{10 \times 5}{6}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
\(\mathrm{v} =\frac{25}{3} \mathrm{~m} / \mathrm{sec}\)
At point \(\mathrm{C}\) we draw velocity component. In vertical direction-
\(\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\)
here
\(\mathrm{h}=10\)
\(\mathrm{u}=\mathrm{v} \sin \theta=\frac{25}{3} \times 0.6\)
Then,
\(10=\frac{25}{3} \times 0.6 \mathrm{t}+\frac{1}{2} \times 10 \times \mathrm{t}^{2}\)
\(10=5 \mathrm{t}+5 \mathrm{t}^{2}\)
\(\mathrm{t}^{2}+\mathrm{t}-2=0\)
\((\mathrm{t}+2)(\mathrm{t}-1)=0\)
\(\mathrm{t}=1 \sec\)
\(\mathrm{DF}=(\) Horizontal component of velocity \() \times\) time
\(\mathrm{DF}=\mathrm{v} \cos \theta \times \mathrm{t}\)
\(=\frac{25}{3} \times 0.8 \times 1\)
\(\mathrm{DF}=\frac{20}{3} \mathrm{~m}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146158 Two touching blocks 1 and 2 are placed on an inclined plane forming an angle \(60^{\circ}\) with the horizontal. The masses are \(m_{1}\) and \(m_{2}\) and the coefficient of friction between the inclined plane and the two blocks are \(1.5 \mu\) and \(1.0 \mu\), respectively. The force of reaction between the blocks during the motion is \((\mathrm{g}=\) acceleration due to gravity)

1 \(\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mu \mathrm{g}\)

2 \(\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mu \mathrm{g}\)

3 \(\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

4 \(\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

Explanation:

D Given,
Coefficient of friction for first block \(\left(\mu_{1}\right)=1.5 \mu\)
Coefficient of friction for second block \(\left(\mu_{2}\right)=1 \mu\)
Acceleration of both block \(=\mathrm{a}\)

Now, FBD for the first block,

If a be the common acceleration, then
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-\mu_{1} \cdot \mathrm{R}_{1}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \sin 60^{\circ}-1.5 \mu \cdot \mathrm{m}_{1} \mathrm{~g} \cos 60^{\circ}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{m}_{1} \mathrm{~g} \frac{\sqrt{3}}{2}-1.5 \mu \mathrm{m}_{1} \mathrm{~g} \times \frac{1}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}=\mathrm{m}_{1} \mathrm{a}\)
Similarly, for second block,

Since, \(\quad \mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu_{2} \mathrm{R}_{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\mathrm{m}_{2} \mathrm{~g} \sin 60^{\circ}-\mu \mathrm{m}_{2} \mathrm{~g} \cos 60^{\circ}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\mu \mathrm{m}_{2} \mathrm{~g} \times \frac{1}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
\(\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}=\mathrm{m}_{2} \mathrm{a}\)
Dividing equation (i) by (ii), we get -
\(\frac{\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}}{2}-\frac{1.5 \mu \mathrm{m}_{1} \mathrm{~g}}{2}+\mathrm{R}}{\frac{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}}{2}-\frac{\mu \mathrm{m}_{2} \mathrm{~g}}{2}-\mathrm{R}}=\frac{\mathrm{m}_{1} \mathrm{a}}{\mathrm{m}_{2} \mathrm{a}}\)
\(\frac{\sqrt{3} \mathrm{~m}_{1} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~g}+2 \mathrm{R}}{\sqrt{3} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{2} \mathrm{~g}-2 \mathrm{R}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\)
\(\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-1.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}+2 \mathrm{Rm}_{2}\)
\(=\sqrt{3} \mathrm{~m}_{1} \mathrm{~m}_{2} \mathrm{~g}-\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}-2 \mathrm{Rm}_{1}\)
\(2 \mathrm{R}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=0.5 \mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}\)
\(\mathrm{R}=\frac{\mu \mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}}{4\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\)
\(\mathrm{R}=\frac{1}{4} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}} \mu \mathrm{g}\)

TS- EAMCET-03.05.2019

Laws of Motion

146159 Consider a system of blocks \(X, A\) and \(B\) as shown in the figure. The blocks \(A\) and \(B\) have equal mass and are connected by a mass-less string through a mass-less pulley. The coefficient of friction between block \(A\) and \(X\) or \(B\) and \(X\) is 0.5. If block \(X\) moves on the horizontal frictionless surface what should be its minimum acceleration such that blocks \(A\) and \(B\) remain stationary, \((g=\) acceleration due to gravity.)

1 \(\frac{g}{3}\)

2 \(3 \mathrm{~g}\)

3 \(\frac{g}{4}\)

4 \(\frac{3 g}{4}\)

TS- EAMCET-03.05.2019

Laws of Motion

146160 A block of mass \(10 \mathrm{~kg}\), initially at rest, makes a downward motion on \(45^{\circ}\) inclined plane. Then the distance travelled by the block after \(2 \mathrm{~s}\) is (Assume the coefficient of kinetic friction to be 0.3 and \(g=10 \mathrm{~ms}^{-2}\) )

1 \(7 \sqrt{2} \mathrm{~m}\)

2 \(\frac{9}{\sqrt{2}} \mathrm{~m}\)

3 \(10 \sqrt{2} \mathrm{~m}\)

4 \(5 \sqrt{2} \mathrm{~m}\)

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane \(\mathrm{BCE}\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular wooden block ABCD of height \(10 \mathrm{~m}\), as shown in the figure. \(A\) small block is allowed to slide down from the top \(E\) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{8}\) and the angle of inclination of the inclined plane is \(\sin ^{-1}(0.6)\). If the small block finally reaches the ground at a point \(F\), then DF will be
(Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{5}{3} \mathrm{~m}\)

2 \(\frac{10}{3} \mathrm{~m}\)

3 \(\frac{13}{3} \mathrm{~m}\)

4 \(\frac{20}{3} \mathrm{~m}\)

Explanation:

D Angle of inclination \(\theta=\sin ^{-1}(0.6)\) \(\sin \theta=0.6\)
\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-(0.6)^{2}}\)
\(\cos \theta=0.8\)

Given, \(\mu_{\mathrm{k}}=\frac{1}{8}\)
Mass of block \(=m\)
Let, acceleration of mass \(=\mathrm{a}\)
Block slide down therefore
Net force on block (downward)
\(\mathrm{R}=\mathrm{mg} \cos \theta\)
\(\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma} \quad\) (downward)
\(\mathrm{mg} \sin \theta-\mu_{\mathrm{k}} \mathrm{mg} \cos \theta=\mathrm{ma}\)
\((\therefore \mathrm{f}=\mu \mathrm{R})\)
\(\mathrm{g} \sin \theta-\mu_{\mathrm{k}} \mathrm{g} \cos \theta=\mathrm{a}\)
\(10 \times 0.6-\frac{1}{8} \times 10 \times 0.8=a\)
\(6-1=a\)
\(a=5 \mathrm{~m} / \mathrm{sec}^{2}\)
Velocity of mass at point \(\mathrm{C}\)
\(v^{2} =u^{2}+2 a s\)
\(v^{2} =0+2 \times 5 \times s\)
\(\mathrm{v}^{2}=10 \mathrm{~s}(\therefore \mathrm{u}=0, \text { velocity at } \mathrm{E}=0)\)
Here, \(\mathrm{s}=\mathrm{EC}\)
In triangle \(\mathrm{ECB}\)
\(\sin \theta=\frac{\mathrm{BE}}{\mathrm{EC}},\left(\mathrm{BE}=\frac{25}{6}\right)\)
\(0.6=\frac{\frac{25}{6}}{\mathrm{EC}}\)
\(\mathrm{EC} =\frac{25}{3.6}\)
\(\therefore \quad \mathrm{v}^{2} =10 \mathrm{~s}\)
\(\mathrm{v}^{2} =10 \times \frac{25}{3.6}=10 \times \frac{25}{36} \times 10\)
\(\mathrm{v} =\frac{10 \times 5}{6}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
\(\mathrm{v} =\frac{25}{3} \mathrm{~m} / \mathrm{sec}\)
At point \(\mathrm{C}\) we draw velocity component. In vertical direction-
\(\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\)
here
\(\mathrm{h}=10\)
\(\mathrm{u}=\mathrm{v} \sin \theta=\frac{25}{3} \times 0.6\)
Then,
\(10=\frac{25}{3} \times 0.6 \mathrm{t}+\frac{1}{2} \times 10 \times \mathrm{t}^{2}\)
\(10=5 \mathrm{t}+5 \mathrm{t}^{2}\)
\(\mathrm{t}^{2}+\mathrm{t}-2=0\)
\((\mathrm{t}+2)(\mathrm{t}-1)=0\)
\(\mathrm{t}=1 \sec\)
\(\mathrm{DF}=(\) Horizontal component of velocity \() \times\) time
\(\mathrm{DF}=\mathrm{v} \cos \theta \times \mathrm{t}\)
\(=\frac{25}{3} \times 0.8 \times 1\)
\(\mathrm{DF}=\frac{20}{3} \mathrm{~m}\)

AP EAMCET (20.04.2019) Shift-1