
A force $\vec F = - K(y\hat i + x\hat j)$ where $K$ is a positive constant, 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 y-axis to the point $(a,a)$. The total work done by the force $F$ on the particle is:
A. \[ - 2K{a^2}\]
B. \[2K{a^2}\]
C. \[ - K{a^2}\]
D. \[K{a^2}\]
Answer
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Hint: Work is the product of the force in the direction of the displacement and the magnitude of this displacement. The SI unit of work is Joule and it is denoted by ‘J’. A force of 1N in moving an object through a distance of 1 meter in the direction of force. Work done is elaborated in such a way that it includes both force extended on the body and total displacement of the body.
Complete step by step answer:
Given, $\vec F = - K(y\hat i + x\hat j)$.......…(1)
While moving from $(0,0)$ to $(a,0)$
Along positive x-axis, $y = 0$
Then
$\vec F = - K(0\hat i + x\hat j)$
$ \Rightarrow \vec F = - Kx\hat j$
i.e, force is in negative y-direction while displacement is in positive x-direction.
$\therefore {W_1} = 0$
Because force is perpendicular to displacement. Then particle moves from $(a,0)$ to $(a,a)$ along a line parallel to y-axis $(x = + a)$ during this
$\vec F = - K(y\hat i + a\hat j)$
The first component of force, $ - ky\hat i$ will not contribute any work because the component is along negative x-direction $( - \hat i)$ while displacement is in positive y-direction $(a,0)$ to $(a,a)$.
The second component of force i.e, $ - ka\hat j$ will perform negative work.
${W_2} = ( - ka\hat j)(a\hat j)$
$ \Rightarrow {W_2} = ( - ka)(a)$
$ \Rightarrow {W_2} = - k{a^2}$
So net work done on the particle
$W = {W_1} + {W_2}$
\[ \Rightarrow W = 0 + ( - k{a^2})\]
\[ \therefore W = - k{a^2}\]
So the correct answer is option C.
Note: A vector is an element of vector space. A vector is an element which has both a direction and the magnitude. For example- force, velocity, displacement. A scalar quantity of an element which possesses the magnitude but not the direction. For example- density, temperature, pressure.
Complete step by step answer:
Given, $\vec F = - K(y\hat i + x\hat j)$.......…(1)
While moving from $(0,0)$ to $(a,0)$
Along positive x-axis, $y = 0$
Then
$\vec F = - K(0\hat i + x\hat j)$
$ \Rightarrow \vec F = - Kx\hat j$
i.e, force is in negative y-direction while displacement is in positive x-direction.
$\therefore {W_1} = 0$
Because force is perpendicular to displacement. Then particle moves from $(a,0)$ to $(a,a)$ along a line parallel to y-axis $(x = + a)$ during this
$\vec F = - K(y\hat i + a\hat j)$
The first component of force, $ - ky\hat i$ will not contribute any work because the component is along negative x-direction $( - \hat i)$ while displacement is in positive y-direction $(a,0)$ to $(a,a)$.
The second component of force i.e, $ - ka\hat j$ will perform negative work.
${W_2} = ( - ka\hat j)(a\hat j)$
$ \Rightarrow {W_2} = ( - ka)(a)$
$ \Rightarrow {W_2} = - k{a^2}$
So net work done on the particle
$W = {W_1} + {W_2}$
\[ \Rightarrow W = 0 + ( - k{a^2})\]
\[ \therefore W = - k{a^2}\]
So the correct answer is option C.
Note: A vector is an element of vector space. A vector is an element which has both a direction and the magnitude. For example- force, velocity, displacement. A scalar quantity of an element which possesses the magnitude but not the direction. For example- density, temperature, pressure.
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