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Newton’s Laws of Motion — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsNewton’s Laws of Motion5 Marks
Question
In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement m1 = 300g and m2 = 600g. The system is released from rest.
  1. Find the distance travelled by the first block in the first two seconds.
  2. Find the tension in the string.
  3. Find the force exerted by the clamp on the pulley.

✓

Answer


m1 = 0.3kg, m2 = 0.6kg

T - (m1g + m1a) = 0 …(i)

⇒ T = m1g + m1a

T + m2a - m2g = 0 …(ii)

⇒ T = m2g - m2a

From equation (i) and equation (ii)

m1g + m1a + m2a - m2g = 0, from (i)

⇒ a(m1 + m2) = g(m2 - m1)

$\Rightarrow\text{a = f}\Big(\frac{\text{m}_2-\text{m}_1}{\text{m}_1-\text{m}_2}\Big)=9.8\Big(\frac{0.6-0.3}{0.6+0.3}\Big)=3.266\text{ms}^{-2}.$

  1. t = 2 sec acceleration = 3.266 ms-2

Initial velocity u = 0

So, distance travelled by the body is,

$\text{S = ut}+\frac{1}{2}\text{at}^2\Rightarrow0+\frac{1}{2}(3.266)2^2=6.5\text{m}$

  1. From (i) T = m1(g + a) = 0.3(9.8 + 3.26) = 3.9N
  2. The force exerted by the clamp on the pully is given by

F - 2T = 0

F = 2T = 2 × 3.9 = 7.8N.

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