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Laws of Motion

270155 A horizontal force $F$ pushes a $4 kg$ block (A) which pushes against a $2 kg$ block (B) as shown. The blocks have an acceleration of $3 m / s^{2}$ to the right. There is no friction between the blocks and the surfaces on which they slide. What is the net force $B$ exerts on $A$ ?

1

6 N

to the right

2

12 N

to the right

3

6 N

to the left

4

12 N

to the left

Explanation:

$F = m a, F_{n e t} = F - f, f = m a$
$a = \frac{T_{3}}{m_{1} + m_{2} + m_{3}}, T_{2} = (m_{1} + m_{2}) a$

Laws of Motion

270156 Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance $S$ as shown in Figure. If $m_{1} < m_{2}$ then the reading of spring balance will be

1

(m_{1} - m_{2})

2

(m_{1} + m_{2})

3

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

4

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

Explanation:

$F_{net} = m a$ ,From FBD of $m_{1}, m_{1} g - T_{1} = m_{1} a$ From FBD of $m_{2}, T_{2} - m_{2} g = m_{2} a$ take $T_{1} = T_{2}$ , solving the above eq's,we get ' $a$ '

Laws of Motion

270157 Two masses $(M + m)$ and $(M - m)$ are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

1

g m / M

2

2 gM / m

3 gm/2M

4

g (M^{2} - m^{2}) / 2 M

Explanation:

$a = g \frac{◻ m_{2} - m_{1}}{◻ m_{1} + m_{2}} ◻$

Laws of Motion

270158 Two bodies of masses $5 kg$ and $4 kg$ are tied to a string as shown. If the table and pulley are smooth, then acceleration of $5 kg$ mass will be

1

19.5 m / s^{2}

2

0.55 m / s^{2}

3

2.72 m / s^{2}

4

5.45 m / s^{2}

Explanation:

$F = m a, a = \frac{m_{2} g}{m_{1} + m_{2}}$

Laws of Motion

270155 A horizontal force $F$ pushes a $4 kg$ block (A) which pushes against a $2 kg$ block (B) as shown. The blocks have an acceleration of $3 m / s^{2}$ to the right. There is no friction between the blocks and the surfaces on which they slide. What is the net force $B$ exerts on $A$ ?

1

6 N

to the right

2

12 N

to the right

3

6 N

to the left

4

12 N

to the left

Explanation:

$F = m a, F_{n e t} = F - f, f = m a$
$a = \frac{T_{3}}{m_{1} + m_{2} + m_{3}}, T_{2} = (m_{1} + m_{2}) a$

Laws of Motion

270156 Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance $S$ as shown in Figure. If $m_{1} < m_{2}$ then the reading of spring balance will be

1

(m_{1} - m_{2})

2

(m_{1} + m_{2})

3

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

4

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

Explanation:

$F_{net} = m a$ ,From FBD of $m_{1}, m_{1} g - T_{1} = m_{1} a$ From FBD of $m_{2}, T_{2} - m_{2} g = m_{2} a$ take $T_{1} = T_{2}$ , solving the above eq's,we get ' $a$ '

Laws of Motion

270157 Two masses $(M + m)$ and $(M - m)$ are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

1

g m / M

2

2 gM / m

3 gm/2M

4

g (M^{2} - m^{2}) / 2 M

Explanation:

$a = g \frac{◻ m_{2} - m_{1}}{◻ m_{1} + m_{2}} ◻$

Laws of Motion

270158 Two bodies of masses $5 kg$ and $4 kg$ are tied to a string as shown. If the table and pulley are smooth, then acceleration of $5 kg$ mass will be

1

19.5 m / s^{2}

2

0.55 m / s^{2}

3

2.72 m / s^{2}

4

5.45 m / s^{2}

Explanation:

$F = m a, a = \frac{m_{2} g}{m_{1} + m_{2}}$

Laws of Motion

270155 A horizontal force $F$ pushes a $4 kg$ block (A) which pushes against a $2 kg$ block (B) as shown. The blocks have an acceleration of $3 m / s^{2}$ to the right. There is no friction between the blocks and the surfaces on which they slide. What is the net force $B$ exerts on $A$ ?

1

6 N

to the right

2

12 N

to the right

3

6 N

to the left

4

12 N

to the left

Explanation:

$F = m a, F_{n e t} = F - f, f = m a$
$a = \frac{T_{3}}{m_{1} + m_{2} + m_{3}}, T_{2} = (m_{1} + m_{2}) a$

Laws of Motion

270156 Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance $S$ as shown in Figure. If $m_{1} < m_{2}$ then the reading of spring balance will be

1

(m_{1} - m_{2})

2

(m_{1} + m_{2})

3

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

4

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

Explanation:

$F_{net} = m a$ ,From FBD of $m_{1}, m_{1} g - T_{1} = m_{1} a$ From FBD of $m_{2}, T_{2} - m_{2} g = m_{2} a$ take $T_{1} = T_{2}$ , solving the above eq's,we get ' $a$ '

Laws of Motion

270157 Two masses $(M + m)$ and $(M - m)$ are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

1

g m / M

2

2 gM / m

3 gm/2M

4

g (M^{2} - m^{2}) / 2 M

Explanation:

$a = g \frac{◻ m_{2} - m_{1}}{◻ m_{1} + m_{2}} ◻$

Laws of Motion

270158 Two bodies of masses $5 kg$ and $4 kg$ are tied to a string as shown. If the table and pulley are smooth, then acceleration of $5 kg$ mass will be

1

19.5 m / s^{2}

2

0.55 m / s^{2}

3

2.72 m / s^{2}

4

5.45 m / s^{2}

Explanation:

$F = m a, a = \frac{m_{2} g}{m_{1} + m_{2}}$

Laws of Motion

270155 A horizontal force $F$ pushes a $4 kg$ block (A) which pushes against a $2 kg$ block (B) as shown. The blocks have an acceleration of $3 m / s^{2}$ to the right. There is no friction between the blocks and the surfaces on which they slide. What is the net force $B$ exerts on $A$ ?

1

6 N

to the right

2

12 N

to the right

3

6 N

to the left

4

12 N

to the left

Explanation:

$F = m a, F_{n e t} = F - f, f = m a$
$a = \frac{T_{3}}{m_{1} + m_{2} + m_{3}}, T_{2} = (m_{1} + m_{2}) a$

Laws of Motion

270156 Two masses $m_{1}$ and $m_{2}$ are attached to a spring balance $S$ as shown in Figure. If $m_{1} < m_{2}$ then the reading of spring balance will be

1

(m_{1} - m_{2})

2

(m_{1} + m_{2})

3

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

4

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

Explanation:

$F_{net} = m a$ ,From FBD of $m_{1}, m_{1} g - T_{1} = m_{1} a$ From FBD of $m_{2}, T_{2} - m_{2} g = m_{2} a$ take $T_{1} = T_{2}$ , solving the above eq's,we get ' $a$ '

Laws of Motion

270157 Two masses $(M + m)$ and $(M - m)$ are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

1

g m / M

2

2 gM / m

3 gm/2M

4

g (M^{2} - m^{2}) / 2 M

Explanation:

$a = g \frac{◻ m_{2} - m_{1}}{◻ m_{1} + m_{2}} ◻$

Laws of Motion

270158 Two bodies of masses $5 kg$ and $4 kg$ are tied to a string as shown. If the table and pulley are smooth, then acceleration of $5 kg$ mass will be

1

19.5 m / s^{2}

2

0.55 m / s^{2}

3

2.72 m / s^{2}

4

5.45 m / s^{2}

Explanation:

$F = m a, a = \frac{m_{2} g}{m_{1} + m_{2}}$