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LAWS OF MOTION (ADDITIONAL)

371727 An object of mass $15 kg$ moves at a constant speed of $15 {ms}^{- 1}$ . A constant force, which acts for 5 seconds on the object gives it a speed 5 ${ms}^{- 1}$ in opposite direction. The force acting on the object is?

1

- 50 N

2

60 N

3

- 40 N

4

- 60 N

Explanation:

D Given, $v_{1} = 15 m / \sec$
$v_{2} = - 5 m / \sec (opposite direction)$
Time $(t) = 5 \sec$
Mass $(m) = 15 kg$
We know force is equal to rate of change of momentum.
$\vec{F} = m \frac{d \vec{v}}{dt}$
$\vec{F} = 15 \times \frac{[(- 5) - 15]}{5}$
$\vec{F} = 15 \times \frac{- 20}{5}$
$\vec{F} = - 60 N$

TS EAMCET (Medical) 09.08.2021

LAWS OF MOTION (ADDITIONAL)

371728 A car moving at a speed ' $v$ ' is stopped by a retarding force ' $F$ ' in a distance ' $s$ '. If the retarding force were $3 F$ . The car will be stopped in a distance

1

\frac{s}{3}

2

\frac{s}{6}

3

\frac{s}{9}

4

\frac{s}{12}

Explanation:

A Let the mass of car be $m$
Retardation $(a) = \frac{- F}{m}$
Now using $v^{2} - u^{2} = 2 a$ s, here final velocity is zero,
We get,
$- u^{2} = 2 \frac{- F}{m} s$
$u^{2} = \frac{2 Fs}{m} \Rightarrow v^{2} = \frac{2 Fs}{m}$
Now the force be 3F
Let the distance be $s_{1}$
Retardation $(a) = \frac{- 3 F}{m}$ velocity is $v$ , final velocity,
$v^{2} = 2 \frac{3 F}{m} s_{1}$
From (i) \& (ii)
$\frac{2 Fs}{m} = 2 \frac{3 F}{m} s_{1}$
$\Rightarrow s_{1} = \frac{s}{3}$

AP EAMCET-24.08.2021

LAWS OF MOTION (ADDITIONAL)

371729 Two blocks $A$ and $B$ of masses $4 kg$ and $6 kg$ are as shown in the figure. $A$ horizontal force of $12 N$ is required to make A slip over B. Find the maximum horizontal force $F_{B}$ that can be applied on $B$ so that both $A$ and $B$ move together $($ take $g = 10 m . s^{- 2})$

1

30 N

2

27 N

3

32 N

4

25 N

Explanation:

A Given that-
$m_{A} = 4 kg, f_{A} = 12 N$
$m_{B} = 6 kg$
$f_{A} = ma \Rightarrow a = \frac{f}{m_{A}} = \frac{12}{4} \Rightarrow a = 3 m / s^{2}$
Now, when blocks move together the acceleration will be same-
Then, $F_{B} - f_{A} = m_{B} a$
$F_{B} - 12 = 6 \times 3$
$F_{B} = 18 + 12$
$F_{B} = 30 N$

AP EAMCET-19.08.2021

LAWS OF MOTION (ADDITIONAL)

371730 An object is moving with a constant speed along a straight-line path. A force is not required to

1 increase its speed

2 decrease its momentum

3 change its direction

4 keep it moving with uniform velocity

AP EAMCET-23.08.2021

LAWS OF MOTION (ADDITIONAL)

371727 An object of mass $15 kg$ moves at a constant speed of $15 {ms}^{- 1}$ . A constant force, which acts for 5 seconds on the object gives it a speed 5 ${ms}^{- 1}$ in opposite direction. The force acting on the object is?

1

- 50 N

2

60 N

3

- 40 N

4

- 60 N

Explanation:

D Given, $v_{1} = 15 m / \sec$
$v_{2} = - 5 m / \sec (opposite direction)$
Time $(t) = 5 \sec$
Mass $(m) = 15 kg$
We know force is equal to rate of change of momentum.
$\vec{F} = m \frac{d \vec{v}}{dt}$
$\vec{F} = 15 \times \frac{[(- 5) - 15]}{5}$
$\vec{F} = 15 \times \frac{- 20}{5}$
$\vec{F} = - 60 N$

TS EAMCET (Medical) 09.08.2021

LAWS OF MOTION (ADDITIONAL)

371728 A car moving at a speed ' $v$ ' is stopped by a retarding force ' $F$ ' in a distance ' $s$ '. If the retarding force were $3 F$ . The car will be stopped in a distance

1

\frac{s}{3}

2

\frac{s}{6}

3

\frac{s}{9}

4

\frac{s}{12}

Explanation:

A Let the mass of car be $m$
Retardation $(a) = \frac{- F}{m}$
Now using $v^{2} - u^{2} = 2 a$ s, here final velocity is zero,
We get,
$- u^{2} = 2 \frac{- F}{m} s$
$u^{2} = \frac{2 Fs}{m} \Rightarrow v^{2} = \frac{2 Fs}{m}$
Now the force be 3F
Let the distance be $s_{1}$
Retardation $(a) = \frac{- 3 F}{m}$ velocity is $v$ , final velocity,
$v^{2} = 2 \frac{3 F}{m} s_{1}$
From (i) \& (ii)
$\frac{2 Fs}{m} = 2 \frac{3 F}{m} s_{1}$
$\Rightarrow s_{1} = \frac{s}{3}$

AP EAMCET-24.08.2021

LAWS OF MOTION (ADDITIONAL)

371729 Two blocks $A$ and $B$ of masses $4 kg$ and $6 kg$ are as shown in the figure. $A$ horizontal force of $12 N$ is required to make A slip over B. Find the maximum horizontal force $F_{B}$ that can be applied on $B$ so that both $A$ and $B$ move together $($ take $g = 10 m . s^{- 2})$

1

30 N

2

27 N

3

32 N

4

25 N

Explanation:

A Given that-
$m_{A} = 4 kg, f_{A} = 12 N$
$m_{B} = 6 kg$
$f_{A} = ma \Rightarrow a = \frac{f}{m_{A}} = \frac{12}{4} \Rightarrow a = 3 m / s^{2}$
Now, when blocks move together the acceleration will be same-
Then, $F_{B} - f_{A} = m_{B} a$
$F_{B} - 12 = 6 \times 3$
$F_{B} = 18 + 12$
$F_{B} = 30 N$

AP EAMCET-19.08.2021

LAWS OF MOTION (ADDITIONAL)

371730 An object is moving with a constant speed along a straight-line path. A force is not required to

1 increase its speed

2 decrease its momentum

3 change its direction

4 keep it moving with uniform velocity

AP EAMCET-23.08.2021

LAWS OF MOTION (ADDITIONAL)

371727 An object of mass $15 kg$ moves at a constant speed of $15 {ms}^{- 1}$ . A constant force, which acts for 5 seconds on the object gives it a speed 5 ${ms}^{- 1}$ in opposite direction. The force acting on the object is?

1

- 50 N

2

60 N

3

- 40 N

4

- 60 N

Explanation:

D Given, $v_{1} = 15 m / \sec$
$v_{2} = - 5 m / \sec (opposite direction)$
Time $(t) = 5 \sec$
Mass $(m) = 15 kg$
We know force is equal to rate of change of momentum.
$\vec{F} = m \frac{d \vec{v}}{dt}$
$\vec{F} = 15 \times \frac{[(- 5) - 15]}{5}$
$\vec{F} = 15 \times \frac{- 20}{5}$
$\vec{F} = - 60 N$

TS EAMCET (Medical) 09.08.2021

LAWS OF MOTION (ADDITIONAL)

371728 A car moving at a speed ' $v$ ' is stopped by a retarding force ' $F$ ' in a distance ' $s$ '. If the retarding force were $3 F$ . The car will be stopped in a distance

1

\frac{s}{3}

2

\frac{s}{6}

3

\frac{s}{9}

4

\frac{s}{12}

Explanation:

A Let the mass of car be $m$
Retardation $(a) = \frac{- F}{m}$
Now using $v^{2} - u^{2} = 2 a$ s, here final velocity is zero,
We get,
$- u^{2} = 2 \frac{- F}{m} s$
$u^{2} = \frac{2 Fs}{m} \Rightarrow v^{2} = \frac{2 Fs}{m}$
Now the force be 3F
Let the distance be $s_{1}$
Retardation $(a) = \frac{- 3 F}{m}$ velocity is $v$ , final velocity,
$v^{2} = 2 \frac{3 F}{m} s_{1}$
From (i) \& (ii)
$\frac{2 Fs}{m} = 2 \frac{3 F}{m} s_{1}$
$\Rightarrow s_{1} = \frac{s}{3}$

AP EAMCET-24.08.2021

LAWS OF MOTION (ADDITIONAL)

371729 Two blocks $A$ and $B$ of masses $4 kg$ and $6 kg$ are as shown in the figure. $A$ horizontal force of $12 N$ is required to make A slip over B. Find the maximum horizontal force $F_{B}$ that can be applied on $B$ so that both $A$ and $B$ move together $($ take $g = 10 m . s^{- 2})$

1

30 N

2

27 N

3

32 N

4

25 N

Explanation:

A Given that-
$m_{A} = 4 kg, f_{A} = 12 N$
$m_{B} = 6 kg$
$f_{A} = ma \Rightarrow a = \frac{f}{m_{A}} = \frac{12}{4} \Rightarrow a = 3 m / s^{2}$
Now, when blocks move together the acceleration will be same-
Then, $F_{B} - f_{A} = m_{B} a$
$F_{B} - 12 = 6 \times 3$
$F_{B} = 18 + 12$
$F_{B} = 30 N$

AP EAMCET-19.08.2021

LAWS OF MOTION (ADDITIONAL)

371730 An object is moving with a constant speed along a straight-line path. A force is not required to

1 increase its speed

2 decrease its momentum

3 change its direction

4 keep it moving with uniform velocity

AP EAMCET-23.08.2021

LAWS OF MOTION (ADDITIONAL)

371727 An object of mass $15 kg$ moves at a constant speed of $15 {ms}^{- 1}$ . A constant force, which acts for 5 seconds on the object gives it a speed 5 ${ms}^{- 1}$ in opposite direction. The force acting on the object is?

1

- 50 N

2

60 N

3

- 40 N

4

- 60 N

Explanation:

D Given, $v_{1} = 15 m / \sec$
$v_{2} = - 5 m / \sec (opposite direction)$
Time $(t) = 5 \sec$
Mass $(m) = 15 kg$
We know force is equal to rate of change of momentum.
$\vec{F} = m \frac{d \vec{v}}{dt}$
$\vec{F} = 15 \times \frac{[(- 5) - 15]}{5}$
$\vec{F} = 15 \times \frac{- 20}{5}$
$\vec{F} = - 60 N$

TS EAMCET (Medical) 09.08.2021

LAWS OF MOTION (ADDITIONAL)

371728 A car moving at a speed ' $v$ ' is stopped by a retarding force ' $F$ ' in a distance ' $s$ '. If the retarding force were $3 F$ . The car will be stopped in a distance

1

\frac{s}{3}

2

\frac{s}{6}

3

\frac{s}{9}

4

\frac{s}{12}

Explanation:

A Let the mass of car be $m$
Retardation $(a) = \frac{- F}{m}$
Now using $v^{2} - u^{2} = 2 a$ s, here final velocity is zero,
We get,
$- u^{2} = 2 \frac{- F}{m} s$
$u^{2} = \frac{2 Fs}{m} \Rightarrow v^{2} = \frac{2 Fs}{m}$
Now the force be 3F
Let the distance be $s_{1}$
Retardation $(a) = \frac{- 3 F}{m}$ velocity is $v$ , final velocity,
$v^{2} = 2 \frac{3 F}{m} s_{1}$
From (i) \& (ii)
$\frac{2 Fs}{m} = 2 \frac{3 F}{m} s_{1}$
$\Rightarrow s_{1} = \frac{s}{3}$

AP EAMCET-24.08.2021

LAWS OF MOTION (ADDITIONAL)

371729 Two blocks $A$ and $B$ of masses $4 kg$ and $6 kg$ are as shown in the figure. $A$ horizontal force of $12 N$ is required to make A slip over B. Find the maximum horizontal force $F_{B}$ that can be applied on $B$ so that both $A$ and $B$ move together $($ take $g = 10 m . s^{- 2})$

1

30 N

2

27 N

3

32 N

4

25 N

Explanation:

A Given that-
$m_{A} = 4 kg, f_{A} = 12 N$
$m_{B} = 6 kg$
$f_{A} = ma \Rightarrow a = \frac{f}{m_{A}} = \frac{12}{4} \Rightarrow a = 3 m / s^{2}$
Now, when blocks move together the acceleration will be same-
Then, $F_{B} - f_{A} = m_{B} a$
$F_{B} - 12 = 6 \times 3$
$F_{B} = 18 + 12$
$F_{B} = 30 N$

AP EAMCET-19.08.2021

LAWS OF MOTION (ADDITIONAL)

371730 An object is moving with a constant speed along a straight-line path. A force is not required to

1 increase its speed

2 decrease its momentum

3 change its direction

4 keep it moving with uniform velocity

AP EAMCET-23.08.2021