SATHEE: Laws Of Motion Question 260

Laws Of Motion Question 260

Question: A body of mass 8kg is moved by a force $F = 3 x N,$ where $x$ is the distance covered. Initial position is $x = 2 m$ and the final position is $x = 10$ m. The initial speed is $0.0 m / s .$ The final speed is[Orissa JEE 2002]

Options:

A) 6 m/s

B) 12 m/s

C) 18 m/s

D) 14 m/s

Show Answer

Answer:

Correct Answer: A

Solution:

Increment in kinetic energy = work done therefore

$\frac{1}{2} m (v^{2} - u^{2}) = \int_{x_{1}}^{x_{2}} F . d x = \int_{2}^{10} (3 x) d x$

therefore $\frac{1}{2} m v^{2} = \frac{3}{2} [x^{2}]_{2}^{10} = \frac{3}{2} [100 - 4]$

therefore $\frac{1}{2} \times 8 \times v^{2} = \frac{3}{2} \times 96$

therefore $v = 6 m / s$

पिछला

अगला

Laws Of Motion Question 260

Question: A body of mass 8kg is moved by a force $F = 3 x N,$ where $x$ is the distance covered. Initial position is $x = 2 m$ and the final position is $x = 10$ m. The initial speed is $0.0 m / s .$ The final speed is[Orissa JEE 2002]

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

Laws Of Motion Question 260

Question: A body of mass 8kg is moved by a force F=3xN, where x is the distance covered. Initial position is x=2m and the final position is x=10 m. The initial speed is 0.0m/s. The final speed is[Orissa JEE 2002]

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 body of mass 8kg is moved by a force $F = 3 x N,$ where $x$ is the distance covered. Initial position is $x = 2 m$ and the final position is $x = 10$ m. The initial speed is $0.0 m / s .$ The final speed is[Orissa JEE 2002]