Consider a force $\vec{F}=-x \hat{i}+y \hat{j}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is: (all quantities are in SI units)

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Work, Power & Energy

Consider a force

F = - x \hat{i} + y \hat{j}

. The work done by this force in moving a particle from point

A (1, 0)

B (0, 1)

along the line segment is: (all quantities are in SI units)

(A)2
(B) $\frac{1}{2}$
(C)1✓ Correct
(D) $\frac{3}{2}$

Answer: (C)

Solution

The work is

\int F \cdot d s = \int (- x d x + y d y)

, and because each force component depends only on its own coordinate the two pieces integrate separately. As

x

runs from

1

0

\int_{1}^{0} - x d x = \frac{1}{2}

, and as

y

runs from

0

1

\int_{0}^{1} y d y = \frac{1}{2}

. Their sum is

1

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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