Work done by a force- F -6 x 3 - N- in displacing a particle from x -4 m to x -2 m is-A- -360 JB- -240 JC- 360 JD- 240 J

The correct option is C 360J W = ∫^ 2_+4 Fdx W = ∫^ 2_+4 6x^3 dx = ∫^4_2 6x^3 dx = 6 [x^4/4]^4_ 2 = 6/4 [x^4]^4_ 2 = 6/4 [(4)^4 (2)^4] = 6/4[256 16] ...

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Standard XII

Physics

Electric Field

Work done by ...

Question

Work done by a force, $F = (- 6 x^{3})^i N$ , in displacing a particle from x = 4m to x = –2m is:

360J

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240J

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–240J

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–360J

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Solution

The correct option is A
360J

$W = \int_{+ 4}^{- 2} F d x W = \int_{+ 4}^{- 2} - 6 x^{3} d x = \int_{2}^{4} 6 x^{3} d x = 6 [\frac{x^{4}}{4}]_{- 2}^{4} = \frac{6}{4} [x^{4}]_{- 2}^{4} = \frac{6}{4} [(4)^{4} - (2)^{4}] = \frac{6}{4} [256 - 16] = \frac{6}{4} \times 240 \Rightarrow 360 J$

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Work done by a force,F=(−6x3)^iN , in displacing a particle from x = 4m to x = –2m is:

The correct option is A 360J W=∫−2+4FdxW=∫−2+4−6x3dx=∫426x3dx=6[x44]4−2=64[x4]4−2=64[(4)4−(2)4]=64[256−16]=64×240⇒360J

Work done by a force, $F = (- 6 x^{3})^i N$ , in displacing a particle from x = 4m to x = –2m is:

The correct option is A
360J

$W = \int_{+ 4}^{- 2} F d x W = \int_{+ 4}^{- 2} - 6 x^{3} d x = \int_{2}^{4} 6 x^{3} d x = 6 [\frac{x^{4}}{4}]_{- 2}^{4} = \frac{6}{4} [x^{4}]_{- 2}^{4} = \frac{6}{4} [(4)^{4} - (2)^{4}] = \frac{6}{4} [256 - 16] = \frac{6}{4} \times 240 \Rightarrow 360 J$