Question

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light string that passes over an ideal pulley. The tension force exerted on block A after the system is released from rest has magnitude T. Block B is then replaced by a block of mass 2M and the system again released from rest.
Which of the following is a correct expression for the tension force now exerted on block A?

a) 12T b) 23T c) 34T d) 43T e) 2T

Answer 1

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

Answer 2

The tension exerted on block A when the mass of block B is replaced is $\frac{4}{3} T$.

The given parameters;

• mass of block A = m
• mass of block B = m
• the tension of block A = T
• new mass of block B = 2m

The initial tension exerted on block A is calculated as follows;

the block is moving horizontally while the block B is pulling the string downwards.

$T = m_A a$

The resultant force on the block B;

$W_B - T = m_Ba$

Solving the two equations together;

$W_B - m_Aa = m_B a\\\\m_Bg - m_Aa = m_B a\\\\m_Bg = m_Ba + m_A a\\\\m_Bg = a(m_B + m_A)\\\\a = \frac{m_Bg}{m_B + m_A} \\\\a = \frac{gm}{m+ m} \\\\a = \frac{gm}{2m} \\\\a = \frac{g}{2}$

The tension on the block A;

$T = m_A a\\\\T = m_A \times \frac{g}{2} \\\\T = \frac{1}{2} m_Ag$

when the mass of block is replaced by 2m

$a = \frac{m_Bg}{m_A + m_B} \\\\a = \frac{2mg}{m + 2m} \\\\a = \frac{2mg}{3m} \\\\a = \frac{2g}{3}$

The new tension on block A is calculated as follows;

$T_2 = m_A a\\\\T_2 = ma \times \frac{2g}{3} \\\\T_2 = \frac{2}{3} m_A g\\\\T_2 = \frac{2}{3} \times \frac{2}{2} \times m_Ag\\\\T_2 = \frac{2\times 2}{3} \times \frac{1}{2} m_A g\\\\T_2 = \frac{4}{3} (\frac{1}{2} m_A g)\\\\T_2 = \frac{4}{3} (T)$

Thus, the tension exerted on block A when the mass of block B is replaced is $\frac{4}{3} T$.

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