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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Physics
Chapter
Work, Power and Energy

BITSAT 2010 · Physics · Work, Power and Energy

The potential energy for a force field F is given by U(x, y)= (x+y)...

The potential energy for a force field F→\overrightarrow{\mathrm{F}} is given by U(x,y)=cos⁡(x+y)U(x, y)=\cos (x+y). The force acting on a particle at position given by coordinates (0,π/4)(0, \pi / 4) is −-

Options

  1. A−12(i^+j^)-\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})
  2. B12(i^+j^)\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})
  3. C(12i^+32j^)\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)
  4. D(12i^−32j^)\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)

Correct answer

B. 12(i^+j^)\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})

Step-by-step solution

Fx=−∂U∂x=sin⁡(x+y)\quad \mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) Fy=−∂U∂x=sin⁡(x+y)Fx=sin⁡(x+y)](0,π/4)=12 Fy=sin⁡(x+y)](0,π/4)=12∴F=12[i^+j^]\begin{array}{l}\mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\\ \left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}}\\ \left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}}\\ \therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}] \end{array}
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