SATHEE: Work Energy And Power Question 227

Work Energy And Power Question 227

Question: A block of mass $m = 0.1$ kg is connected to a spring of unknown spring constant k. It is compressed to a distance $(\frac{x}{2})$ from its equilibrium position and released from rest. After approaching half the distance from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity $3 m s^{- 1}$ . The total initial energy of the spring is

Options:

A) $0.3 J$

B) $0.6 J$

C) $0.8 J$

D) $1.5 J$

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Answer:

Correct Answer: B

Solution:

[b] Applying momentum conservation $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$0.1 u + m (0) = 0.1 (0) + m (3)$

$0.1 u = 3 m$

$\frac{1}{2} 0.1 u^{2} = \frac{1}{2} m {(3)}^{2}$ Solving we get, $u = 3$

$\frac{1}{2} k x^{2} = \frac{1}{2} K {(\frac{x}{2})}^{2} + \frac{1}{2} (0.1) 3^{2}$

$\Rightarrow \frac{3}{4} k x^{2} = 0.9 \Rightarrow \frac{3}{2} \times \frac{1}{2} k x^{2} = 0.9$

$∴ \frac{1}{2} K x^{2} = 0.6 J$ (total initial energy of the spring)

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Work Energy And Power Question 227

Options:

Answer:

Solution:

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Work Energy And Power Question 227

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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