Question

The only force acting on a $2.0kg$ body as it moves along a positive $x$ -axis has an $x$ -component $F_x=-6x$ with $x$ in metres. The velocity at $x=3.0m$ is $8.0m/s$ . The velocity of the body at $x=4.0m$ is

• A. $6.6m/s$
• B. $46.6m/s$
• C. $60m/s$
• D. $96.6m/s$

Answer 1

Answer: (A)

As the body moves along the $x$-axis from
$x_i=3.0m$ to $x_f=4.0m$, the work done by the force is
$W=\underset{x_i}{\overset{x_f}{\int }}{f}_{x}dx=\underset{x_i}{\overset{x_f}{\int }}-6xdx=-3\left(x_f^2-x_i^2\right)$
$=-3\left[{\left(4.0\right)}^2-{\left(3.0\right)}^2\right]11=-21J$
According to the work-kinetic energy theorem, this gives the change in the kinetic energy.
$W=\Delta K=\frac{1}{2}m(v_f^2-v_i^2)$
where, $v_i$ is the initial velocity (at $x_i$ and $v_f$ is the final velocity (at $x_f$). The theorem yields.
$v_f=\sqrt{\frac{2W}{m}+v_i^2}$
$=\sqrt{\frac{2\left(-21\right)}{2.0}+{\left(8.0\right)}^2}$
$=6.6m/s$