A particle moves in a force field F = (y^2)i + (2xy)j. Calculate the work done in moving the particle from (0,0) to (1,1) along the path y = x^2.

A. 1 — The force field is conservative because the curl of F is zero. Therefore, the work done is independent of the path and can be calculated using the potential function U where F = -grad(U).

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$A particle moves in a force field F = y^{2} \hat{i} + 2 x y \hat{j} . Calculate the work done in moving the particle from (0, 0) to (1, 1) along the path y = x^{2} .$

A. $1$
B. $0.5$
C. $2$
D. $0$

Show correct answer & step-by-step solution

Correct answer: A — $1$

Solution

The force field is conservative since \nabla \times F = 0 . The potential function is U = - x y^{2}, so W = - Δ U = U (0, 0) - U (1, 1) = 0 - (- 1) = 1 .

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A particle moves in a force field F=y2i^+2xyj^​. Calculate the work done in moving the particle from (0,0) to (1,1) along the path y=x2.

Solution

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More Work Energy Power practice questions

$A particle moves in a force field F = y^{2} \hat{i} + 2 x y \hat{j} . Calculate the work done in moving the particle from (0, 0) to (1, 1) along the path y = x^{2} .$