SATHEE: Work Energy And Power Question 233

Work Energy And Power Question 233

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

$W = \int \vec{F} . \vec{d l}$

In going from (0,0) to (a, 0), the coordinate ofx varies from 0 to ‘a’, while that of .y remains zero.

Hence, the work done along this path is: $W_{1} = \int_{0}^{a} (- k x \hat{j}) . d x \hat{i} = 0 [∵ \hat{j} . \hat{i} = 0]$

In going from $(a, 0) t o (a, a)$ the coordinate of-c remains constant $(= a)$

while that of y changes from 0 to ‘a’. Hence, the work done along this path is $W_{2} = \int_{0}^{a} [- K (y \hat{i} + a \hat{j}) . d y \hat{j}] = k a \int_{0}^{a} d y = - k a^{2}$ Hence, $W = W_{1} + W_{2} = - k a^{2}$

Work Energy And Power Question 233

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Solution:

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Work Energy And Power Question 233

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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