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Work Done on & by the Object

A force of ...

Question

A force of $\to F = 2 x^i + 2^j + 3 z^{2}^k$ N is acting on a particle. Find the work done by this force in displacing the body from (1, 2, 3)m to (3, 6, 1) m.

14 J

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4100 J

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1.4 J

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1 J

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Solution

The correct option is A 14 J
$\to F = 2 x^i + 2^j + 3 z^{2}^k$
$\Rightarrow_{x} = 2 x$ $_{y} = 2$ $_{2} = 3 z^{2}$
Work done, $W = \int \to F . \to d r$
$\Rightarrow W = \int \to F . \to d r = \int_{1}^{3} 2 x . d x + \int_{2}^{6} 2 y . d y + \int_{3}^{1} 3 z^{2} . d x$
$= {[x^{2}]}_{1}^{3} + {[y^{2}]}_{2}^{6} + {[z^{3}]}_{3}^{1}$
$= 8 + 32 - 26$
$= 14 J .$
Hence, the answer is $14 J .$

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A force of →F=2x^i+2^j+3z2^k N is acting on a particle. Find the work done by this force in displacing the body from (1, 2, 3)m to (3, 6, 1) m.

The correct option is A 14 J→F=2x^i+2^j+3z2^k⇒→Fx=2x →Fy=2 →F2=3z2Work done, W=∫→F.→dr⇒W=∫→F.→dr=∫312x.dx+∫622y.dy+∫133z2.dx =[x2]31+[y2]62+[z3]13 =8+32−26 =14J.Hence, the answer is 14J.

A force of $\to F = 2 x^i + 2^j + 3 z^{2}^k$ N is acting on a particle. Find the work done by this force in displacing the body from (1, 2, 3)m to (3, 6, 1) m.