A force of $(4 x^{2} + 3 x) N$ acts on a particle which displaces it from $x = 2 m$ to $x = 3 m$ . The work done by the force is:

32.8 J

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3.28 J

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0.328 J

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z e r o

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Solution

The correct option is A $32.8 J$
$W = \int F d x = \int (4 x^{2} + 3 x) d x = \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2}$
$W = \frac{4 \times 3^{3}}{3} + \frac{3 \times 3^{2}}{2} - \frac{4 \times 2^{3}}{3} - \frac{3 \times 2^{2}}{2} = 32.8 J$

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Similar questions

(4 x^{2} + 3 x) N

x = 2 m

x = 3 m

F = (^i + 3^j + 4^k) N

(- 1 m, 2 m, 1 m)

(2 m, - 3 m, 1 m)

{\underset{r}{\to}}_{1} = (2 m) \underset{i}{\to} + (3 m) \underset{j}{\to}

{\underset{r}{\to}}_{2} = (3 m) \underset{i}{\to} + (2 m) \underset{j}{\to}

\to F = (- 6 x^{3}^i) N

x = 4 m

x = - 2 m

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