Kinematics Question 114

Question: A force F=K(yi^+xj^) (where K it’s 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 forces F on the particle is [IIT-JEE 1998]

Options:

A) 2Ka2

B) 2Ka2

C) Ka2

D) Ka2

Show Answer

Answer:

Correct Answer: C

Solution:

For motion of the particle from (0, 0) to (a, 0)

F=K(0i^+aj^)
F=Kaj^

Displacement r=(ai^+0j^)(0i^+0j^)=ai^

So work done from (0, 0) to (a, 0) it’s given by W=F.r

=Kaj^.ai^=0

For motion (a, 0) to (a, a) F=K(ai^+aj^) and displacement r=(ai^+aj^)(ai^+0j^)=aj^

So work done from (a, 0) to (a, a) W=F.r

=K(ai^+aj^).aj^=Ka2

So total work done =Ka2



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