Work Energy And Power Question 65

Question: A force F=K(yi+xj) (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 [IIT 1998]

Options:

A) 2Ka2

B) 2Ka2

C) Ka2

D) Ka2

Show Answer

Answer:

Correct Answer: C

Solution:

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0
 F=kxj^ 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(yi^+aJ^)

The first component of force, kyi^ will not contribute any work because this component 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. kaj^ will perform negative work \ W2=(kaj^) (aj^) = (ka) (a) =ka2

So net work done on the particle W=W1+W2 = 0+(ka2)=ka2



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