Question
A force F=−K(yi+aj) (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 particles is
Answer: Option C
:
C
While moving from (0,0) to (a,0)
Along positive x-axis, y = 0 ∴→F=−kx^j
i.e. force is in negative y-direction while displacement is in positive x-direction.
∴W1=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→F=−k(y^i+a^j)
The first component of force, −ky^i will not contribute any work because this components is along negative x-direction (−^i)while displacement is in positive y-direction (a,0) to (a,a).
The second component of force i.e. −ka^j
∴W2=(−ka^j)(a^j)=(−ka)(a)=−ka2
So net work done on the particle W=W1+W2
=0+(−ka2)=−ka2
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:
C
While moving from (0,0) to (a,0)
Along positive x-axis, y = 0 ∴→F=−kx^j
i.e. force is in negative y-direction while displacement is in positive x-direction.
∴W1=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→F=−k(y^i+a^j)
The first component of force, −ky^i will not contribute any work because this components is along negative x-direction (−^i)while displacement is in positive y-direction (a,0) to (a,a).
The second component of force i.e. −ka^j
∴W2=(−ka^j)(a^j)=(−ka)(a)=−ka2
So net work done on the particle W=W1+W2
=0+(−ka2)=−ka2
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