SATHEE: Work Energy And Power Question 208

Work Energy And Power Question 208

Question: Calculate the work done on the tool by $\vec{F} (11.25 \hat{i} + 11.25 \hat{j}) N$ if the tool is first moved out along the x-axis to the point $x = 3.00 m,$

$y = 0$ and then moved parallel to the y-axis to $x = 3.00 m,$

$y = 3.00 m$ .

Options:

A) 67.5 J

B) 85 J

C) 102 J

D) 7.5 J

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Net displacement is $(3 \hat{i} + 3 \hat{j})$

so work done $= \vec{F} . Δ \vec{s} = 11.25 (\hat{i} + \hat{j}) . (3 \hat{i} + 3 \hat{j})$

$= 33.75 \times 2 = 67.50 J$

पिछला

अगला

Work Energy And Power Question 208

Question: Calculate the work done on the tool by $\vec{F} (11.25 \hat{i} + 11.25 \hat{j}) N$ if the tool is first moved out along the x-axis to the point $x = 3.00 m,$

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 208

Question: Calculate the work done on the tool by F→(11.25i^+11.25j^)N if the tool is first moved out along the x-axis to the point x=3.00m,

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

Menu

Question: Calculate the work done on the tool by $\vec{F} (11.25 \hat{i} + 11.25 \hat{j}) N$ if the tool is first moved out along the x-axis to the point $x = 3.00 m,$