SATHEE: Work Energy And Power Question 308

Work Energy And Power Question 308

Question: A force $\vec{F} = - k (y \hat{i} + x \hat{j})$ acts on a particle moving in the x -y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the j’-axis to the point (a, a). The total work done by the force is

Options:

A) $- 2 k a^{2}$

B) $2 k a^{2}$

C) $- k a^{2}$

D) $k a^{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $W_{1} \int_{0}^{a} \vec{F} . \vec{d x} = \int_{0}^{a} - k (y \hat{i} - x \hat{j}) . \hat{i} d x$

$= \int_{0}^{a} - k (0 \hat{i} + x \hat{j}) . \hat{i} d x = z e r o$

$W \int_{0}^{a} \vec{F} . \vec{d y} = \int_{0}^{a} - k (y \hat{i} + x \hat{j}) . \hat{j} d y$

$= \int_{0}^{a} - k (a \hat{i} + a \hat{j}) . \hat{j} d y$

$= - k a \int_{0}^{a} d y = - k a^{2}$

Total work done, $W = W_{1} + W_{2} = 0 - k a^{2} = - k a^{2}$

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Work Energy And Power Question 308

Question: A force F→=−k(yi^+xj^) acts on a particle moving in the x -y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the j’-axis to the point (a, a). The total work done by the force is

Options:

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Question: A force $\vec{F} = - k (y \hat{i} + x \hat{j})$ acts on a particle moving in the x -y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the j’-axis to the point (a, a). The total work done by the force is