Respuesta :
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,
[tex]F(x)=(3\ N/m^2)x^2+(6\ N/m)x[/tex]
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 :
[tex]W=\int\limits {F{\cdot} dx}[/tex]
[tex]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[/tex]
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Answer:
Explanation:
Force is given by
F = 3x² + 6x
Particle moves from x = 0 m to 2 m
Work done is
[tex]W=\int_{x_{1}}^{x_{2}}F(x)dx[/tex]
[tex]W=\int_{0}^{2}\left ( 3x^{2}+6x \right )dx[/tex]
[tex]W=\left ( x^{3}+3x^{2} \right )_{0}^{2}[/tex]
W = 8 + 12 - 0 - 0
W = 20 J
