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NEET 2005
Exam
NEET
Year
2005
Subject
Physics
Chapter
Work, Power and Energy

NEET 2005 · Physics · Work, Power and Energy

A force F acting on an object varies with distance x as shown here....

A force FF acting on an object varies with distance xx as shown here. The force is in NN and xx in m\mathrm{m}. The work done by the force in moving the object x=0x=0 to x=6 mx=6 \mathrm{~m} is: A force F acting on an object varies with distance x as shown here. The force is in N and x in m . The work done by the

Options

  1. A18.0 J18.0 \mathrm{~J}
  2. B13.5 J13.5 \mathrm{~J}
  3. C9.0 J9.0 \mathrm{~J}
  4. D4.5 J4.5 \mathrm{~J}

Correct answer

B. 13.5 J13.5 \mathrm{~J}

Step-by-step solution

Work done = Area under =f−x=f-x curve. = area of trapezium =12×(6+3)×3=272=13.5 J\begin{aligned}& =\text{ area of trapezium }=\frac{1}{2} \times(6+3) \times3 \\ & =\frac{27}{2}=13.5 \mathrm{~J} \end{aligned}
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