Question

F = K (xi+y) is working on a particle which moves from (0, 0) to (a,0) along x axis and
(a,0) to ſa. a) along y axis. Find the work done by the force.
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Answer 1

Answer:

Given:

$force = k(x \hat{i } + y \hat{j})$

Object moves from (0,0) to (a,0) along x axis and (a,0) to (a,a) along y axis.

To find:

Work done by the force

Concept:

Work done is the Product of Force with displacement. It is the amount of energy given by the force to move the body along the displacement.

Calculation:

Since the displacement here is a function of x and y , we need to perform integration separately.

$\sf{ \int dw = k \{( \int x \: dx )+ ( \int y dy) \}}$

Now putting the limits :

$\sf{ \implies\int_{0}^{w} \: dw = k \{( \int_{0}^{a} x \: dx )+ ( \int_{0}^{a} y dy) \}}$

$\sf{ \implies \: w = k \bigg \{ \frac{ {x}^{2} }{2} \bigg \}_{0}^{a} + k \bigg \{ \frac{ {y}^{2} }{2} \bigg \}_{0}^{a} }$

$\sf{ \implies \: w = \frac{1}{2}k {a}^{2} + \frac{1}{2}k {a}^{2} }$

$\sf{ \implies \: w = k {a}^{2} }$

So final answer :

$\boxed{ \huge{ \blue{ \bold{ \sf{ \: w = k {a}^{2} }}}}}$

Answer 2

$\huge\red{\boxed{\boxed{\boxed{\boxed{\boxed{\mathbb{\blue{\underline{\underline{ w = ka^2}}}}}}}}}}$