SATHEE: Work Energy And Power Question 63

Work Energy And Power Question 63

Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $- K / r^{2}$ , where K is a constant. The total energy of the particle is [IIT 1977]

Options:

A) $\frac{K}{2 r}$

B) $- \frac{K}{2 r}$

C) $- \frac{K}{r}$

D) $\frac{K}{r}$

Show Answer

Answer:

Correct Answer: B

Solution:

Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}}$

K.E. $= \frac{1}{2} m v^{2} = \frac{K}{2 r}$

$U = - \int_{\infty}^{r} F . d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$

Total energy $E = K .E . + P .E . = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$

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Work Energy And Power Question 63

Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $- K / r^{2}$ , where K is a constant. The total energy of the particle is [IIT 1977]

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

Work Energy And Power Question 63

Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to −K/r2 , where K is a constant. The total energy of the particle is [IIT 1977]

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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Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $- K / r^{2}$ , where K is a constant. The total energy of the particle is [IIT 1977]