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Laws of Motion

146094 An object takes $n$ times as much time as to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down perfectly smooth inclined planned of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by

1

(1 - \frac{1}{n^{2}})

2

(\frac{1}{1 - n^{2}})

3

\sqrt{1 - \frac{1}{n^{2}}}

4

\sqrt{1 + \frac{1}{n^{2}}}

Explanation:

A We know that,
Smooth plane of acceleration (a) $= g \sin θ$
For rough plane of acceleration $(a) = g (\sin θ - μ \cos θ)$
From, Newton's Law of motion,
$s = ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} {at}^{2} (∵ u = 0)$
$t = \sqrt{\frac{2 s}{a}}$
$t_{rough} = {nt}_{smooth}$
$\sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \times \sqrt{\frac{2 s}{g \sin θ}}$
Squaring both side, we get-
$\frac{2 s}{g (\sin θ - μ \cos θ)} = n^{2} \times \frac{2 s}{g \sin θ}$
$n^{2} = \frac{g \sin θ}{g (\sin θ - μ \cos θ)}$
$n^{2} = \frac{\sin 45^{\circ}}{\sin 45^{\circ} - μ \cos 45^{\circ}} (∵ θ = 45^{\circ})$
$n^{2} = \frac{1 / \sqrt{2}}{1 / \sqrt{2} - μ \times 1 / \sqrt{2}}$
$n^{2} = \frac{1}{(1 - μ)}$
$n^{2} - n^{2} μ = 1$
$n^{2} μ = n^{2} - 1$
$μ = (1 - \frac{1}{n^{2}})$

AP EAMCET-24.04.2019

Laws of Motion

146095 A block of mass $m$ is in contact with the cart $C$ as shown in the figure.

The coefficient of static friction between the block and the cart is $μ$ . The acceleration $α$ of the cart that will prevent the block from falling satisfies

1

α > \frac{m g}{μ}

2

α > \frac{g}{μ m}

3

α \geq \frac{g}{μ}

4

α < \frac{g}{μ}

Explanation:

C
The above diagram display as the forces working on a block of mass $m$ .
Therefore,
$m α =$ pseudo forces
Frictional forces
$F_{s} \geq mg$
$μ N \geq mg (∴ N = m α)$
$μ m α \geq mg$
$\Rightarrow α \geq \frac{g}{μ}$

AIPMT- 2010

Laws of Motion

146097 A body of mass $m$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 m$ after reaching the bottom. What is the value of $h$ ?
[Use $g = 10 m / s^{2}$ ]

1

2 m

2

\frac{5}{2} m

3

5 m

4

10 m

Explanation:

C
The body can describe circular path if the velocity of the body at lowest point $B$ of the circular path is-
$v = \sqrt{5 gr}$
$v = \sqrt{5 \times 10 \times 2}$
$v = \sqrt{100} = 10 m / s$
At point $A$ there is only potential energy and at the point B energy is completely kinetic
$∴$ Kinetic energy of body at point B is -
$K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times m \times (10)^{2} = 50 m$
$∴$ Potential energy at point $A$ is-
$= mgh$
From law of conservation of energy-
$P.E. = K.E.$
$mgh = 50 m$
$h = \frac{50}{g} = 5 m$

TS EAMCET 19.07.2022

Laws of Motion

146098 A cyclist is riding with a speed of $36 km / h$ . As he approaches a circular turn on the road of radius 50, he applies brakes and reduces his speed at the constant rate of $0.5 m / s$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are

1

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (4)

2

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (\frac{1}{4})

3

\sqrt{17} m / s^{2}, \tan^{- 1} (\frac{1}{4})

4

\frac{\sqrt{17}}{2} m / s^{2}, \tan^{- 1} (4)

Explanation:

D Radius of the circular turn $(r) = 50 m$
Tangential acceleration $({\vec{a}}_{t}) = 0.5 m / s^{2}$
Centripetal acceleration $({\vec{a}}_{c}) =$ ?

We know that,
${\vec{a}}_{c} = \frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2 m / s^{2}$
As shown in figure two accelerations are multually perpendicular directions.
So, the resultant acceleration is given by
${\vec{a}}_{net} = \sqrt{{\vec{a}}_{t}^{2} + {\vec{a}}_{c}^{2}} = \sqrt{(0.5)^{2} + (2)^{2}}$
${\vec{a}}_{net} = \frac{\sqrt{17}}{2} m / s^{2}$
Hence, $\tan θ = \frac{a_{c}}{a_{t}} = \frac{2}{0.5} = 4$
$θ = \tan^{- 1} (4)$

TS EAMCET 20.07.2022

Laws of Motion

146099 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

Laws of Motion

146094 An object takes $n$ times as much time as to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down perfectly smooth inclined planned of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by

1

(1 - \frac{1}{n^{2}})

2

(\frac{1}{1 - n^{2}})

3

\sqrt{1 - \frac{1}{n^{2}}}

4

\sqrt{1 + \frac{1}{n^{2}}}

Explanation:

A We know that,
Smooth plane of acceleration (a) $= g \sin θ$
For rough plane of acceleration $(a) = g (\sin θ - μ \cos θ)$
From, Newton's Law of motion,
$s = ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} {at}^{2} (∵ u = 0)$
$t = \sqrt{\frac{2 s}{a}}$
$t_{rough} = {nt}_{smooth}$
$\sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \times \sqrt{\frac{2 s}{g \sin θ}}$
Squaring both side, we get-
$\frac{2 s}{g (\sin θ - μ \cos θ)} = n^{2} \times \frac{2 s}{g \sin θ}$
$n^{2} = \frac{g \sin θ}{g (\sin θ - μ \cos θ)}$
$n^{2} = \frac{\sin 45^{\circ}}{\sin 45^{\circ} - μ \cos 45^{\circ}} (∵ θ = 45^{\circ})$
$n^{2} = \frac{1 / \sqrt{2}}{1 / \sqrt{2} - μ \times 1 / \sqrt{2}}$
$n^{2} = \frac{1}{(1 - μ)}$
$n^{2} - n^{2} μ = 1$
$n^{2} μ = n^{2} - 1$
$μ = (1 - \frac{1}{n^{2}})$

AP EAMCET-24.04.2019

Laws of Motion

146095 A block of mass $m$ is in contact with the cart $C$ as shown in the figure.

The coefficient of static friction between the block and the cart is $μ$ . The acceleration $α$ of the cart that will prevent the block from falling satisfies

1

α > \frac{m g}{μ}

2

α > \frac{g}{μ m}

3

α \geq \frac{g}{μ}

4

α < \frac{g}{μ}

Explanation:

C
The above diagram display as the forces working on a block of mass $m$ .
Therefore,
$m α =$ pseudo forces
Frictional forces
$F_{s} \geq mg$
$μ N \geq mg (∴ N = m α)$
$μ m α \geq mg$
$\Rightarrow α \geq \frac{g}{μ}$

AIPMT- 2010

Laws of Motion

146097 A body of mass $m$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 m$ after reaching the bottom. What is the value of $h$ ?
[Use $g = 10 m / s^{2}$ ]

1

2 m

2

\frac{5}{2} m

3

5 m

4

10 m

Explanation:

C
The body can describe circular path if the velocity of the body at lowest point $B$ of the circular path is-
$v = \sqrt{5 gr}$
$v = \sqrt{5 \times 10 \times 2}$
$v = \sqrt{100} = 10 m / s$
At point $A$ there is only potential energy and at the point B energy is completely kinetic
$∴$ Kinetic energy of body at point B is -
$K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times m \times (10)^{2} = 50 m$
$∴$ Potential energy at point $A$ is-
$= mgh$
From law of conservation of energy-
$P.E. = K.E.$
$mgh = 50 m$
$h = \frac{50}{g} = 5 m$

TS EAMCET 19.07.2022

Laws of Motion

146098 A cyclist is riding with a speed of $36 km / h$ . As he approaches a circular turn on the road of radius 50, he applies brakes and reduces his speed at the constant rate of $0.5 m / s$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are

1

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (4)

2

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (\frac{1}{4})

3

\sqrt{17} m / s^{2}, \tan^{- 1} (\frac{1}{4})

4

\frac{\sqrt{17}}{2} m / s^{2}, \tan^{- 1} (4)

Explanation:

D Radius of the circular turn $(r) = 50 m$
Tangential acceleration $({\vec{a}}_{t}) = 0.5 m / s^{2}$
Centripetal acceleration $({\vec{a}}_{c}) =$ ?

We know that,
${\vec{a}}_{c} = \frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2 m / s^{2}$
As shown in figure two accelerations are multually perpendicular directions.
So, the resultant acceleration is given by
${\vec{a}}_{net} = \sqrt{{\vec{a}}_{t}^{2} + {\vec{a}}_{c}^{2}} = \sqrt{(0.5)^{2} + (2)^{2}}$
${\vec{a}}_{net} = \frac{\sqrt{17}}{2} m / s^{2}$
Hence, $\tan θ = \frac{a_{c}}{a_{t}} = \frac{2}{0.5} = 4$
$θ = \tan^{- 1} (4)$

TS EAMCET 20.07.2022

Laws of Motion

146099 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

Laws of Motion

146094 An object takes $n$ times as much time as to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down perfectly smooth inclined planned of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by

1

(1 - \frac{1}{n^{2}})

2

(\frac{1}{1 - n^{2}})

3

\sqrt{1 - \frac{1}{n^{2}}}

4

\sqrt{1 + \frac{1}{n^{2}}}

Explanation:

A We know that,
Smooth plane of acceleration (a) $= g \sin θ$
For rough plane of acceleration $(a) = g (\sin θ - μ \cos θ)$
From, Newton's Law of motion,
$s = ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} {at}^{2} (∵ u = 0)$
$t = \sqrt{\frac{2 s}{a}}$
$t_{rough} = {nt}_{smooth}$
$\sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \times \sqrt{\frac{2 s}{g \sin θ}}$
Squaring both side, we get-
$\frac{2 s}{g (\sin θ - μ \cos θ)} = n^{2} \times \frac{2 s}{g \sin θ}$
$n^{2} = \frac{g \sin θ}{g (\sin θ - μ \cos θ)}$
$n^{2} = \frac{\sin 45^{\circ}}{\sin 45^{\circ} - μ \cos 45^{\circ}} (∵ θ = 45^{\circ})$
$n^{2} = \frac{1 / \sqrt{2}}{1 / \sqrt{2} - μ \times 1 / \sqrt{2}}$
$n^{2} = \frac{1}{(1 - μ)}$
$n^{2} - n^{2} μ = 1$
$n^{2} μ = n^{2} - 1$
$μ = (1 - \frac{1}{n^{2}})$

AP EAMCET-24.04.2019

Laws of Motion

146095 A block of mass $m$ is in contact with the cart $C$ as shown in the figure.

The coefficient of static friction between the block and the cart is $μ$ . The acceleration $α$ of the cart that will prevent the block from falling satisfies

1

α > \frac{m g}{μ}

2

α > \frac{g}{μ m}

3

α \geq \frac{g}{μ}

4

α < \frac{g}{μ}

Explanation:

C
The above diagram display as the forces working on a block of mass $m$ .
Therefore,
$m α =$ pseudo forces
Frictional forces
$F_{s} \geq mg$
$μ N \geq mg (∴ N = m α)$
$μ m α \geq mg$
$\Rightarrow α \geq \frac{g}{μ}$

AIPMT- 2010

Laws of Motion

146097 A body of mass $m$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 m$ after reaching the bottom. What is the value of $h$ ?
[Use $g = 10 m / s^{2}$ ]

1

2 m

2

\frac{5}{2} m

3

5 m

4

10 m

Explanation:

C
The body can describe circular path if the velocity of the body at lowest point $B$ of the circular path is-
$v = \sqrt{5 gr}$
$v = \sqrt{5 \times 10 \times 2}$
$v = \sqrt{100} = 10 m / s$
At point $A$ there is only potential energy and at the point B energy is completely kinetic
$∴$ Kinetic energy of body at point B is -
$K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times m \times (10)^{2} = 50 m$
$∴$ Potential energy at point $A$ is-
$= mgh$
From law of conservation of energy-
$P.E. = K.E.$
$mgh = 50 m$
$h = \frac{50}{g} = 5 m$

TS EAMCET 19.07.2022

Laws of Motion

146098 A cyclist is riding with a speed of $36 km / h$ . As he approaches a circular turn on the road of radius 50, he applies brakes and reduces his speed at the constant rate of $0.5 m / s$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are

1

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (4)

2

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (\frac{1}{4})

3

\sqrt{17} m / s^{2}, \tan^{- 1} (\frac{1}{4})

4

\frac{\sqrt{17}}{2} m / s^{2}, \tan^{- 1} (4)

Explanation:

D Radius of the circular turn $(r) = 50 m$
Tangential acceleration $({\vec{a}}_{t}) = 0.5 m / s^{2}$
Centripetal acceleration $({\vec{a}}_{c}) =$ ?

We know that,
${\vec{a}}_{c} = \frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2 m / s^{2}$
As shown in figure two accelerations are multually perpendicular directions.
So, the resultant acceleration is given by
${\vec{a}}_{net} = \sqrt{{\vec{a}}_{t}^{2} + {\vec{a}}_{c}^{2}} = \sqrt{(0.5)^{2} + (2)^{2}}$
${\vec{a}}_{net} = \frac{\sqrt{17}}{2} m / s^{2}$
Hence, $\tan θ = \frac{a_{c}}{a_{t}} = \frac{2}{0.5} = 4$
$θ = \tan^{- 1} (4)$

TS EAMCET 20.07.2022

Laws of Motion

146099 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

Laws of Motion

146094 An object takes $n$ times as much time as to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down perfectly smooth inclined planned of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by

1

(1 - \frac{1}{n^{2}})

2

(\frac{1}{1 - n^{2}})

3

\sqrt{1 - \frac{1}{n^{2}}}

4

\sqrt{1 + \frac{1}{n^{2}}}

Explanation:

A We know that,
Smooth plane of acceleration (a) $= g \sin θ$
For rough plane of acceleration $(a) = g (\sin θ - μ \cos θ)$
From, Newton's Law of motion,
$s = ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} {at}^{2} (∵ u = 0)$
$t = \sqrt{\frac{2 s}{a}}$
$t_{rough} = {nt}_{smooth}$
$\sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \times \sqrt{\frac{2 s}{g \sin θ}}$
Squaring both side, we get-
$\frac{2 s}{g (\sin θ - μ \cos θ)} = n^{2} \times \frac{2 s}{g \sin θ}$
$n^{2} = \frac{g \sin θ}{g (\sin θ - μ \cos θ)}$
$n^{2} = \frac{\sin 45^{\circ}}{\sin 45^{\circ} - μ \cos 45^{\circ}} (∵ θ = 45^{\circ})$
$n^{2} = \frac{1 / \sqrt{2}}{1 / \sqrt{2} - μ \times 1 / \sqrt{2}}$
$n^{2} = \frac{1}{(1 - μ)}$
$n^{2} - n^{2} μ = 1$
$n^{2} μ = n^{2} - 1$
$μ = (1 - \frac{1}{n^{2}})$

AP EAMCET-24.04.2019

Laws of Motion

146095 A block of mass $m$ is in contact with the cart $C$ as shown in the figure.

The coefficient of static friction between the block and the cart is $μ$ . The acceleration $α$ of the cart that will prevent the block from falling satisfies

1

α > \frac{m g}{μ}

2

α > \frac{g}{μ m}

3

α \geq \frac{g}{μ}

4

α < \frac{g}{μ}

Explanation:

C
The above diagram display as the forces working on a block of mass $m$ .
Therefore,
$m α =$ pseudo forces
Frictional forces
$F_{s} \geq mg$
$μ N \geq mg (∴ N = m α)$
$μ m α \geq mg$
$\Rightarrow α \geq \frac{g}{μ}$

AIPMT- 2010

Laws of Motion

146097 A body of mass $m$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 m$ after reaching the bottom. What is the value of $h$ ?
[Use $g = 10 m / s^{2}$ ]

1

2 m

2

\frac{5}{2} m

3

5 m

4

10 m

Explanation:

C
The body can describe circular path if the velocity of the body at lowest point $B$ of the circular path is-
$v = \sqrt{5 gr}$
$v = \sqrt{5 \times 10 \times 2}$
$v = \sqrt{100} = 10 m / s$
At point $A$ there is only potential energy and at the point B energy is completely kinetic
$∴$ Kinetic energy of body at point B is -
$K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times m \times (10)^{2} = 50 m$
$∴$ Potential energy at point $A$ is-
$= mgh$
From law of conservation of energy-
$P.E. = K.E.$
$mgh = 50 m$
$h = \frac{50}{g} = 5 m$

TS EAMCET 19.07.2022

Laws of Motion

146098 A cyclist is riding with a speed of $36 km / h$ . As he approaches a circular turn on the road of radius 50, he applies brakes and reduces his speed at the constant rate of $0.5 m / s$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are

1

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (4)

2

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (\frac{1}{4})

3

\sqrt{17} m / s^{2}, \tan^{- 1} (\frac{1}{4})

4

\frac{\sqrt{17}}{2} m / s^{2}, \tan^{- 1} (4)

Explanation:

D Radius of the circular turn $(r) = 50 m$
Tangential acceleration $({\vec{a}}_{t}) = 0.5 m / s^{2}$
Centripetal acceleration $({\vec{a}}_{c}) =$ ?

We know that,
${\vec{a}}_{c} = \frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2 m / s^{2}$
As shown in figure two accelerations are multually perpendicular directions.
So, the resultant acceleration is given by
${\vec{a}}_{net} = \sqrt{{\vec{a}}_{t}^{2} + {\vec{a}}_{c}^{2}} = \sqrt{(0.5)^{2} + (2)^{2}}$
${\vec{a}}_{net} = \frac{\sqrt{17}}{2} m / s^{2}$
Hence, $\tan θ = \frac{a_{c}}{a_{t}} = \frac{2}{0.5} = 4$
$θ = \tan^{- 1} (4)$

TS EAMCET 20.07.2022

Laws of Motion

146099 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

Laws of Motion

146094 An object takes $n$ times as much time as to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down perfectly smooth inclined planned of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by

1

(1 - \frac{1}{n^{2}})

2

(\frac{1}{1 - n^{2}})

3

\sqrt{1 - \frac{1}{n^{2}}}

4

\sqrt{1 + \frac{1}{n^{2}}}

Explanation:

A We know that,
Smooth plane of acceleration (a) $= g \sin θ$
For rough plane of acceleration $(a) = g (\sin θ - μ \cos θ)$
From, Newton's Law of motion,
$s = ut + \frac{1}{2} {at}^{2} = 0 + \frac{1}{2} {at}^{2} (∵ u = 0)$
$t = \sqrt{\frac{2 s}{a}}$
$t_{rough} = {nt}_{smooth}$
$\sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \times \sqrt{\frac{2 s}{g \sin θ}}$
Squaring both side, we get-
$\frac{2 s}{g (\sin θ - μ \cos θ)} = n^{2} \times \frac{2 s}{g \sin θ}$
$n^{2} = \frac{g \sin θ}{g (\sin θ - μ \cos θ)}$
$n^{2} = \frac{\sin 45^{\circ}}{\sin 45^{\circ} - μ \cos 45^{\circ}} (∵ θ = 45^{\circ})$
$n^{2} = \frac{1 / \sqrt{2}}{1 / \sqrt{2} - μ \times 1 / \sqrt{2}}$
$n^{2} = \frac{1}{(1 - μ)}$
$n^{2} - n^{2} μ = 1$
$n^{2} μ = n^{2} - 1$
$μ = (1 - \frac{1}{n^{2}})$

AP EAMCET-24.04.2019

Laws of Motion

146095 A block of mass $m$ is in contact with the cart $C$ as shown in the figure.

The coefficient of static friction between the block and the cart is $μ$ . The acceleration $α$ of the cart that will prevent the block from falling satisfies

1

α > \frac{m g}{μ}

2

α > \frac{g}{μ m}

3

α \geq \frac{g}{μ}

4

α < \frac{g}{μ}

Explanation:

C
The above diagram display as the forces working on a block of mass $m$ .
Therefore,
$m α =$ pseudo forces
Frictional forces
$F_{s} \geq mg$
$μ N \geq mg (∴ N = m α)$
$μ m α \geq mg$
$\Rightarrow α \geq \frac{g}{μ}$

AIPMT- 2010

Laws of Motion

146097 A body of mass $m$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 m$ after reaching the bottom. What is the value of $h$ ?
[Use $g = 10 m / s^{2}$ ]

1

2 m

2

\frac{5}{2} m

3

5 m

4

10 m

Explanation:

C
The body can describe circular path if the velocity of the body at lowest point $B$ of the circular path is-
$v = \sqrt{5 gr}$
$v = \sqrt{5 \times 10 \times 2}$
$v = \sqrt{100} = 10 m / s$
At point $A$ there is only potential energy and at the point B energy is completely kinetic
$∴$ Kinetic energy of body at point B is -
$K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} \times m \times (10)^{2} = 50 m$
$∴$ Potential energy at point $A$ is-
$= mgh$
From law of conservation of energy-
$P.E. = K.E.$
$mgh = 50 m$
$h = \frac{50}{g} = 5 m$

TS EAMCET 19.07.2022

Laws of Motion

146098 A cyclist is riding with a speed of $36 km / h$ . As he approaches a circular turn on the road of radius 50, he applies brakes and reduces his speed at the constant rate of $0.5 m / s$ every second. The magnitude and direction of the net acceleration of the cyclist on the circular turn respectively are

1

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (4)

2

\frac{\sqrt{3}}{2} m / s^{2}, \tan^{- 1} (\frac{1}{4})

3

\sqrt{17} m / s^{2}, \tan^{- 1} (\frac{1}{4})

4

\frac{\sqrt{17}}{2} m / s^{2}, \tan^{- 1} (4)

Explanation:

D Radius of the circular turn $(r) = 50 m$
Tangential acceleration $({\vec{a}}_{t}) = 0.5 m / s^{2}$
Centripetal acceleration $({\vec{a}}_{c}) =$ ?

We know that,
${\vec{a}}_{c} = \frac{v^{2}}{r} = \frac{(10)^{2}}{50} = 2 m / s^{2}$
As shown in figure two accelerations are multually perpendicular directions.
So, the resultant acceleration is given by
${\vec{a}}_{net} = \sqrt{{\vec{a}}_{t}^{2} + {\vec{a}}_{c}^{2}} = \sqrt{(0.5)^{2} + (2)^{2}}$
${\vec{a}}_{net} = \frac{\sqrt{17}}{2} m / s^{2}$
Hence, $\tan θ = \frac{a_{c}}{a_{t}} = \frac{2}{0.5} = 4$
$θ = \tan^{- 1} (4)$

TS EAMCET 20.07.2022

Laws of Motion

146099 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

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