The only force acting on a 1.8-kg body as it moves along the x axis is given by F_x = −(3.0x) N, where x is in m. If the velocity of the body at x = 0 is v_x = +8.0 m/s, at what value of x will the body have a velocity of +4.0 m/s? A) 5.7 m B) 5.4 m C) 4.8 m D) 4.1 m E) 6.6 m

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Quizplus · Accepted Answer

To find the position $x$ where the velocity of the body is $+4.0 \, \text{m/s}$, we can use the work-energy principle. The work done by the force $F = -(3.0x) \, \text{N}$ as the body moves from $x = 0$ to some position $x$ is equal to the change in kinetic energy of the body.The work done by the force is given by the integral of $F \, dx$ from $0$ to $x$, which is $\int_{0}^{x} -(3.0x) \, dx = -\frac{3}{2}x^2$.The kinetic energy at any point is given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. The change in kinetic energy from $x = 0$ to some position $x$ is $\Delta KE = \frac{1}{2}m(v^2 - v_0^2)$, where $v_0 = +8.0 \, \text{m/s}$ is the initial velocity and $v = +4.0 \, \text{m/s}$ is the final velocity.Setting the work equal to the change in kinetic energy gives:$$-\frac{3}{2}x^2 = \frac{1}{2}(1.8)(4^2 - 8^2)$$Solving for $x^2$ gives:$$x^2 = \frac{(1.8)(4^2 - 8^2)}{-3}$$$$x^2 = \frac{(1.8)(16 - 64)}{-3} = \frac{(1.8)(-48)}{-3} = \frac{-86.4}{-3} = 28.8$$Taking the square root of both sides gives $x = \sqrt{28.8}$, which is approximately $5.4 \, \text{m}$.

The Only Force Acting on a 1