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

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsNewton’s Laws of Motion5 Marks
Question
In a simple Atwood machine, two unequal masses $m_1$ and $m_2$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement $m_1 = 300g$ and $m_2 = 600g$. 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



$m_1 = 0.3\ kg, m_2 = 0.6\ kg$
$T - (m_1g + m_1a) = 0 …(i)$
$\Rightarrow T = m_1g + m_1a$
$T + m_2a - m_2g = 0 …(ii)$
$\Rightarrow T = m_2g - m_2a$
From equation $(i)$ and equation $(ii)$
$m_1g + m_1a + m_2a - m_2g = 0, from (i)$
$\Rightarrow a(m_1 + m_2) = g(m_2 - m_1)$
$\Rightarrow\text{a = f}\Big(\frac{\text{m}_2-\text{m}_1}{\text{m}_1-\text{m}_2}\Big)=9.8\Big(\frac{0.6-0.3}{0.6+0.3}\Big)=3.266\text{ms}^{-2}.$
  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 $(i) \ T = m_1(g + a) = 0.3(9.8 + 3.26) = 3.9N$
  2. The force exerted by the clamp on the pully is given by
$F - 2T = 0$
$F = 2T = 2 \times 3.9 = 7.8N.$

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