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

Question

10 days ago

11

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?

2 answers:

Yuliya22 [1.1K] · Answer 1 · 2026-02-23T20:13:53+03:00

8 0

Response:

Details:

The force acting on the particle is given by

F = 3x² + 6x

The particle transitions from x = 0 m to 2 m

The work accomplished 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

Sav [1.1K] · Answer 2 · 2026-02-23T20:13:38+03:00

Response:

The work performed by the particle traveling from x = 0 to x = 2 m totals 20 J.

Details:

The force impacting a particle, which is restricted to the x-axis, is expressed as follows:

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

We need to calculate the work done on a particle moving from x = 0.00 m to x = 2.00 m.

The formula for the work done by the particle is defined as:

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

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

Consequently, the work executed by the particle between x = 0 and x = 2 m amounts to 20 J. Thus, this is the solution sought.