Question

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along the x-axis. What work is done by this force on a particle that moves from x = 0.00 m to x = 2.00 m?

Answer 1

Answer:

Explanation:

Force is given by

F = 3x² + 6x

Particle moves from x = 0 m to 2 m

Work done is

$W=\int_{x_{1}}^{x_{2}}F(x)dx$

$W=\int_{0}^{2}\left ( 3x^{2}+6x \right )dx$

$W=\left ( x^{3}+3x^{2} \right )_{0}^{2}$

W = 8 + 12 - 0 - 0

W = 20 J

Answer 2

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.