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A force F = kx (where k is positive constant) is acting on a particle. Match column–I and column–II, regarding work done in displacing the particle.

Column  i Column  ii
(a) From x = – 4 to x = – 2 (P) Positive
(b) From x = – 2 to x = – 4 (Q) zero
(c) From x = – 2 to x = + 2 (R) negative

(A) a – R, b – P, c – Q 

(B) a – P, b – Q, c – R 

(C) a – R, b – Q, c – P 

(D) a – Q, b – P, c – R

1 Answer

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Best answer

Correct option: (A) a – R, b – P, c – Q

Explanation:

Given : F = kx 

workdone = w = x(2)x(1) F ∙ dx

w = x(2)x(1) k × dx

w = k[x2 / 2]x(2)x(1) = (k/2) [x22 – x12]

(1) from        x = – 4 to x = – 2,

w = (k/2) [(– 2)2 – (– 4)2] = (k/2) (– 12) = – 6 k = negative

(2) from        x = – 2 to x = – 4

w = (k/2) [(– 4)2 – (–2)2] = 6k = positive

(3) form        x = – 2 to x = + 2

w = (k/2) [(2)2 – (– 2)2] = 0   

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