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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 $1.5 μ$ and $1.0 μ$ , respectively. The force of reaction between the blocks during the motion is $(g =$ acceleration due to gravity)

1

(m_{2} - m_{1}) μ g

2

(m_{2} + m_{1}) μ g

3

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

4

\frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

Explanation:

D Given,
Coefficient of friction for first block $(μ_{1}) = 1.5 μ$
Coefficient of friction for second block $(μ_{2}) = 1 μ$
Acceleration of both block $= a$

Now, FBD for the first block,

If a be the common acceleration, then
$m_{1} g \sin 60^{\circ} - μ_{1} \cdot R_{1} + R = m_{1} a$
$m_{1} g \sin 60^{\circ} - 1.5 μ \cdot m_{1} g \cos 60^{\circ} + R = m_{1} a$
$m_{1} g \frac{\sqrt{3}}{2} - 1.5 μ m_{1} g \times \frac{1}{2} + R = m_{1} a$
$\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R = m_{1} a$
Similarly, for second block,

Since, $m_{2} g \sin 60^{\circ} - μ_{2} R_{2} - R = m_{2} a$
$m_{2} g \sin 60^{\circ} - μ m_{2} g \cos 60^{\circ} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - μ m_{2} g \times \frac{1}{2} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R = m_{2} a$
Dividing equation (i) by (ii), we get -
$\frac{\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R}{\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R} = \frac{m_{1} a}{m_{2} a}$
$\frac{\sqrt{3} m_{1} g - 1.5 μ m_{1} g + 2 R}{\sqrt{3} m_{2} g - μ m_{2} g - 2 R} = \frac{m_{1}}{m_{2}}$
$\sqrt{3} m_{1} m_{2} g - 1.5 μ m_{1} m_{2} g + 2 {Rm}_{2}$
$= \sqrt{3} m_{1} m_{2} g - μ m_{1} m_{2} g - 2 {Rm}_{1}$
$2 R (m_{1} + m_{2}) = 0.5 μ m_{1} m_{2} g$
$R = \frac{μ m_{1} m_{2} g}{4 (m_{1} + m_{2})}$
$R = \frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g$

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 g

3

\frac{g}{4}

4

\frac{3 g}{4}

Explanation:

A Given that, $μ = 0.5$ , mass of block A and B, $m_{A} = m_{B} = m$

According to question, $ma + f_{A} = T$
$ma + μ mg = mg - μ ma$
$m (a + μ g) = m (g - μ a)$
$a + μ g = g - μ a$
$a + (\frac{1}{2}) g = g - (\frac{1}{2}) a$
$a + \frac{a}{2} = g - \frac{g}{2}$
$\frac{2 a + a}{2} = \frac{2 g - g}{2}$
$3 a = g$
$a = \frac{g}{3}$
Hence, for stationary of block A and block B minimum acceleration is $g / 3$ .

TS- EAMCET-03.05.2019

Laws of Motion

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

1

7 \sqrt{2} m

2

\frac{9}{\sqrt{2}} m

3

10 \sqrt{2} m

4

5 \sqrt{2} m

Explanation:

A Given, $m = 10 kg, θ = 45^{\circ}, t = 2 s, u = 0 m / s$

From the figure,
$ma = mg \sin θ - μ mg \cos θ$
$a = g \sin θ - μ g \cos θ$
$a = 10 \frac{1}{\sqrt{2}} - 0.3 \times 10 \times \frac{1}{\sqrt{2}} (∵ θ = 45^{\circ})$
$a = \frac{10}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}} {ms}^{- 2}$
Displacement of the block, from the second equation of the motion,
$s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} {at}^{2} (∵ u = 0)$
$s = \frac{1}{2} \times \frac{7}{\sqrt{2}} \times 2^{2}$
$s = 7 \sqrt{2} m$

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane $BCE$ of height $(\frac{25}{6}) m$ is kept on a rectangular wooden block ABCD of height $10 m$ , 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, $g = 10 {ms}^{- 2}$ )

1

\frac{5}{3} m

2

\frac{10}{3} m

3

\frac{13}{3} m

4

\frac{20}{3} m

Explanation:

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

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

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 $1.5 μ$ and $1.0 μ$ , respectively. The force of reaction between the blocks during the motion is $(g =$ acceleration due to gravity)

1

(m_{2} - m_{1}) μ g

2

(m_{2} + m_{1}) μ g

3

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

4

\frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

Explanation:

D Given,
Coefficient of friction for first block $(μ_{1}) = 1.5 μ$
Coefficient of friction for second block $(μ_{2}) = 1 μ$
Acceleration of both block $= a$

Now, FBD for the first block,

If a be the common acceleration, then
$m_{1} g \sin 60^{\circ} - μ_{1} \cdot R_{1} + R = m_{1} a$
$m_{1} g \sin 60^{\circ} - 1.5 μ \cdot m_{1} g \cos 60^{\circ} + R = m_{1} a$
$m_{1} g \frac{\sqrt{3}}{2} - 1.5 μ m_{1} g \times \frac{1}{2} + R = m_{1} a$
$\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R = m_{1} a$
Similarly, for second block,

Since, $m_{2} g \sin 60^{\circ} - μ_{2} R_{2} - R = m_{2} a$
$m_{2} g \sin 60^{\circ} - μ m_{2} g \cos 60^{\circ} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - μ m_{2} g \times \frac{1}{2} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R = m_{2} a$
Dividing equation (i) by (ii), we get -
$\frac{\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R}{\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R} = \frac{m_{1} a}{m_{2} a}$
$\frac{\sqrt{3} m_{1} g - 1.5 μ m_{1} g + 2 R}{\sqrt{3} m_{2} g - μ m_{2} g - 2 R} = \frac{m_{1}}{m_{2}}$
$\sqrt{3} m_{1} m_{2} g - 1.5 μ m_{1} m_{2} g + 2 {Rm}_{2}$
$= \sqrt{3} m_{1} m_{2} g - μ m_{1} m_{2} g - 2 {Rm}_{1}$
$2 R (m_{1} + m_{2}) = 0.5 μ m_{1} m_{2} g$
$R = \frac{μ m_{1} m_{2} g}{4 (m_{1} + m_{2})}$
$R = \frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g$

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 g

3

\frac{g}{4}

4

\frac{3 g}{4}

Explanation:

A Given that, $μ = 0.5$ , mass of block A and B, $m_{A} = m_{B} = m$

According to question, $ma + f_{A} = T$
$ma + μ mg = mg - μ ma$
$m (a + μ g) = m (g - μ a)$
$a + μ g = g - μ a$
$a + (\frac{1}{2}) g = g - (\frac{1}{2}) a$
$a + \frac{a}{2} = g - \frac{g}{2}$
$\frac{2 a + a}{2} = \frac{2 g - g}{2}$
$3 a = g$
$a = \frac{g}{3}$
Hence, for stationary of block A and block B minimum acceleration is $g / 3$ .

TS- EAMCET-03.05.2019

Laws of Motion

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

1

7 \sqrt{2} m

2

\frac{9}{\sqrt{2}} m

3

10 \sqrt{2} m

4

5 \sqrt{2} m

Explanation:

A Given, $m = 10 kg, θ = 45^{\circ}, t = 2 s, u = 0 m / s$

From the figure,
$ma = mg \sin θ - μ mg \cos θ$
$a = g \sin θ - μ g \cos θ$
$a = 10 \frac{1}{\sqrt{2}} - 0.3 \times 10 \times \frac{1}{\sqrt{2}} (∵ θ = 45^{\circ})$
$a = \frac{10}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}} {ms}^{- 2}$
Displacement of the block, from the second equation of the motion,
$s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} {at}^{2} (∵ u = 0)$
$s = \frac{1}{2} \times \frac{7}{\sqrt{2}} \times 2^{2}$
$s = 7 \sqrt{2} m$

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane $BCE$ of height $(\frac{25}{6}) m$ is kept on a rectangular wooden block ABCD of height $10 m$ , 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, $g = 10 {ms}^{- 2}$ )

1

\frac{5}{3} m

2

\frac{10}{3} m

3

\frac{13}{3} m

4

\frac{20}{3} m

Explanation:

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

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

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 $1.5 μ$ and $1.0 μ$ , respectively. The force of reaction between the blocks during the motion is $(g =$ acceleration due to gravity)

1

(m_{2} - m_{1}) μ g

2

(m_{2} + m_{1}) μ g

3

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

4

\frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

Explanation:

D Given,
Coefficient of friction for first block $(μ_{1}) = 1.5 μ$
Coefficient of friction for second block $(μ_{2}) = 1 μ$
Acceleration of both block $= a$

Now, FBD for the first block,

If a be the common acceleration, then
$m_{1} g \sin 60^{\circ} - μ_{1} \cdot R_{1} + R = m_{1} a$
$m_{1} g \sin 60^{\circ} - 1.5 μ \cdot m_{1} g \cos 60^{\circ} + R = m_{1} a$
$m_{1} g \frac{\sqrt{3}}{2} - 1.5 μ m_{1} g \times \frac{1}{2} + R = m_{1} a$
$\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R = m_{1} a$
Similarly, for second block,

Since, $m_{2} g \sin 60^{\circ} - μ_{2} R_{2} - R = m_{2} a$
$m_{2} g \sin 60^{\circ} - μ m_{2} g \cos 60^{\circ} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - μ m_{2} g \times \frac{1}{2} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R = m_{2} a$
Dividing equation (i) by (ii), we get -
$\frac{\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R}{\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R} = \frac{m_{1} a}{m_{2} a}$
$\frac{\sqrt{3} m_{1} g - 1.5 μ m_{1} g + 2 R}{\sqrt{3} m_{2} g - μ m_{2} g - 2 R} = \frac{m_{1}}{m_{2}}$
$\sqrt{3} m_{1} m_{2} g - 1.5 μ m_{1} m_{2} g + 2 {Rm}_{2}$
$= \sqrt{3} m_{1} m_{2} g - μ m_{1} m_{2} g - 2 {Rm}_{1}$
$2 R (m_{1} + m_{2}) = 0.5 μ m_{1} m_{2} g$
$R = \frac{μ m_{1} m_{2} g}{4 (m_{1} + m_{2})}$
$R = \frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g$

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 g

3

\frac{g}{4}

4

\frac{3 g}{4}

Explanation:

A Given that, $μ = 0.5$ , mass of block A and B, $m_{A} = m_{B} = m$

According to question, $ma + f_{A} = T$
$ma + μ mg = mg - μ ma$
$m (a + μ g) = m (g - μ a)$
$a + μ g = g - μ a$
$a + (\frac{1}{2}) g = g - (\frac{1}{2}) a$
$a + \frac{a}{2} = g - \frac{g}{2}$
$\frac{2 a + a}{2} = \frac{2 g - g}{2}$
$3 a = g$
$a = \frac{g}{3}$
Hence, for stationary of block A and block B minimum acceleration is $g / 3$ .

TS- EAMCET-03.05.2019

Laws of Motion

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

1

7 \sqrt{2} m

2

\frac{9}{\sqrt{2}} m

3

10 \sqrt{2} m

4

5 \sqrt{2} m

Explanation:

A Given, $m = 10 kg, θ = 45^{\circ}, t = 2 s, u = 0 m / s$

From the figure,
$ma = mg \sin θ - μ mg \cos θ$
$a = g \sin θ - μ g \cos θ$
$a = 10 \frac{1}{\sqrt{2}} - 0.3 \times 10 \times \frac{1}{\sqrt{2}} (∵ θ = 45^{\circ})$
$a = \frac{10}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}} {ms}^{- 2}$
Displacement of the block, from the second equation of the motion,
$s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} {at}^{2} (∵ u = 0)$
$s = \frac{1}{2} \times \frac{7}{\sqrt{2}} \times 2^{2}$
$s = 7 \sqrt{2} m$

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane $BCE$ of height $(\frac{25}{6}) m$ is kept on a rectangular wooden block ABCD of height $10 m$ , 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, $g = 10 {ms}^{- 2}$ )

1

\frac{5}{3} m

2

\frac{10}{3} m

3

\frac{13}{3} m

4

\frac{20}{3} m

Explanation:

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

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

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 $1.5 μ$ and $1.0 μ$ , respectively. The force of reaction between the blocks during the motion is $(g =$ acceleration due to gravity)

1

(m_{2} - m_{1}) μ g

2

(m_{2} + m_{1}) μ g

3

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

4

\frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g

Explanation:

D Given,
Coefficient of friction for first block $(μ_{1}) = 1.5 μ$
Coefficient of friction for second block $(μ_{2}) = 1 μ$
Acceleration of both block $= a$

Now, FBD for the first block,

If a be the common acceleration, then
$m_{1} g \sin 60^{\circ} - μ_{1} \cdot R_{1} + R = m_{1} a$
$m_{1} g \sin 60^{\circ} - 1.5 μ \cdot m_{1} g \cos 60^{\circ} + R = m_{1} a$
$m_{1} g \frac{\sqrt{3}}{2} - 1.5 μ m_{1} g \times \frac{1}{2} + R = m_{1} a$
$\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R = m_{1} a$
Similarly, for second block,

Since, $m_{2} g \sin 60^{\circ} - μ_{2} R_{2} - R = m_{2} a$
$m_{2} g \sin 60^{\circ} - μ m_{2} g \cos 60^{\circ} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - μ m_{2} g \times \frac{1}{2} - R = m_{2} a$
$\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R = m_{2} a$
Dividing equation (i) by (ii), we get -
$\frac{\frac{\sqrt{3} m_{1} g}{2} - \frac{1.5 μ m_{1} g}{2} + R}{\frac{\sqrt{3} m_{2} g}{2} - \frac{μ m_{2} g}{2} - R} = \frac{m_{1} a}{m_{2} a}$
$\frac{\sqrt{3} m_{1} g - 1.5 μ m_{1} g + 2 R}{\sqrt{3} m_{2} g - μ m_{2} g - 2 R} = \frac{m_{1}}{m_{2}}$
$\sqrt{3} m_{1} m_{2} g - 1.5 μ m_{1} m_{2} g + 2 {Rm}_{2}$
$= \sqrt{3} m_{1} m_{2} g - μ m_{1} m_{2} g - 2 {Rm}_{1}$
$2 R (m_{1} + m_{2}) = 0.5 μ m_{1} m_{2} g$
$R = \frac{μ m_{1} m_{2} g}{4 (m_{1} + m_{2})}$
$R = \frac{1}{4} \frac{m_{1} m_{2}}{m_{1} + m_{2}} μ g$

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 g

3

\frac{g}{4}

4

\frac{3 g}{4}

Explanation:

A Given that, $μ = 0.5$ , mass of block A and B, $m_{A} = m_{B} = m$

According to question, $ma + f_{A} = T$
$ma + μ mg = mg - μ ma$
$m (a + μ g) = m (g - μ a)$
$a + μ g = g - μ a$
$a + (\frac{1}{2}) g = g - (\frac{1}{2}) a$
$a + \frac{a}{2} = g - \frac{g}{2}$
$\frac{2 a + a}{2} = \frac{2 g - g}{2}$
$3 a = g$
$a = \frac{g}{3}$
Hence, for stationary of block A and block B minimum acceleration is $g / 3$ .

TS- EAMCET-03.05.2019

Laws of Motion

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

1

7 \sqrt{2} m

2

\frac{9}{\sqrt{2}} m

3

10 \sqrt{2} m

4

5 \sqrt{2} m

Explanation:

A Given, $m = 10 kg, θ = 45^{\circ}, t = 2 s, u = 0 m / s$

From the figure,
$ma = mg \sin θ - μ mg \cos θ$
$a = g \sin θ - μ g \cos θ$
$a = 10 \frac{1}{\sqrt{2}} - 0.3 \times 10 \times \frac{1}{\sqrt{2}} (∵ θ = 45^{\circ})$
$a = \frac{10}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}} {ms}^{- 2}$
Displacement of the block, from the second equation of the motion,
$s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} {at}^{2} (∵ u = 0)$
$s = \frac{1}{2} \times \frac{7}{\sqrt{2}} \times 2^{2}$
$s = 7 \sqrt{2} m$

TS- EAMCET-04.05.2019

Laws of Motion

146161 A rough inclined plane $BCE$ of height $(\frac{25}{6}) m$ is kept on a rectangular wooden block ABCD of height $10 m$ , 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, $g = 10 {ms}^{- 2}$ )

1

\frac{5}{3} m

2

\frac{10}{3} m

3

\frac{13}{3} m

4

\frac{20}{3} m

Explanation:

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

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

AP EAMCET (20.04.2019) Shift-1

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