A frictionless pulley has a rope of negligible mass passing over it. Blocks of mass 15 kg and 5 kg are attached at the two ends of the rope and held stationary. When the masses are released to move freely, their speed in m s-1, after the rope has slipped by 0.25 m over the pulley, is closest to :
(take g = 10.0 m s-2)
5.0
2.2
1.6
1.2
Answer is Right!
Answer is Wrong!
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UPSC Geoscientist – 2024
– Let the initial height of the 15 kg mass be $y_1$ and the 5 kg mass be $y_2$. When the rope slips by 0.25 m, the 15 kg mass moves down by 0.25 m (new height $y_1 – 0.25$) and the 5 kg mass moves up by 0.25 m (new height $y_2 + 0.25$).
– The initial potential energy (PE) can be set relative to the initial positions. $\text{PE}_{initial} = m_1 g y_1 + m_2 g y_2$.
– The final potential energy is $\text{PE}_{final} = m_1 g (y_1 – 0.25) + m_2 g (y_2 + 0.25) = m_1 g y_1 – 0.25 m_1 g + m_2 g y_2 + 0.25 m_2 g$.
– The change in potential energy is $\Delta \text{PE} = \text{PE}_{final} – \text{PE}_{initial} = -0.25 m_1 g + 0.25 m_2 g = 0.25 g (m_2 – m_1)$.
– Given $m_1 = 15$ kg, $m_2 = 5$ kg, and $g = 10.0$ m/s², $\Delta \text{PE} = 0.25 \times 10 \times (5 – 15) = 2.5 \times (-10) = -25$ J. The system loses 25 J of potential energy.
– The initial kinetic energy (KE) is zero as the masses are held stationary. $\text{KE}_{initial} = 0$.
– Let $v$ be the speed of the masses after moving 0.25 m. Both masses move with the same speed. The final kinetic energy is $\text{KE}_{final} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 = \frac{1}{2} (m_1 + m_2) v^2$.
– $\text{KE}_{final} = \frac{1}{2} (15 + 5) v^2 = \frac{1}{2} (20) v^2 = 10 v^2$.
– By the conservation of mechanical energy, $\Delta \text{KE} + \Delta \text{PE} = 0$.
– $(10 v^2 – 0) + (-25) = 0$.
– $10 v^2 = 25$.
– $v^2 = \frac{25}{10} = 2.5$.
– $v = \sqrt{2.5}$ m/s.
– Calculating the value: $\sqrt{2.5} \approx 1.581$ m/s.
– Comparing with the options, 1.6 m/s is the closest value.