Q. 27

# In figure (5-E11) m_{1} = 5 kg, m_{2} = kg and F = 1 N. Find the acceleration of either block. Describe the motion of m if the string breaks but F continues to act.

Answer :

Give:

m_{1}= 5 kg, m_{2} = 2 kg and F = 1 N

Let the acceleration of the blocks be a.

From the free-body diagram, m_{1}a = m_{1}g + F − T …(i)

Again, from the free-body diagram, m_{2}a = T − m_{2}g − F …(ii)

Adding equations (i) and (ii), we have:

⇒ a= =4.2 m/s^{2}

Hence, acceleration of the block is 4.2 m/s^{2}.

After the string breaks, m_{1} moves downward with force F acting downward. Then,

m_{1}a = F + m_{1}g

5a = 1 + 5g

g = +0.2 m/s^{2}

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PREVIOUSA constant force F=m2g/2 is applied on the block of mass m1 as shown in figure (5-E 10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.NEXTLet m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5-E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?

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