Answer
At point B, the particle's speed is 3.5 m/s
At point C, the particle's speed is 2.8 m/s
At point D, the particle's speed is 4.5 m/s
Work Step by Step
We know the total energy $E$ is equal to the sum of the potential energy $U$ and the kinetic energy $K$. At point A, the particle is at rest. We can find the total energy.
$E = U+K$
$E = 5.0~J+0$
$E = 5.0~J$
The total energy is 5.0 J
We can find the kinetic energy at point B;
$U+K = E$
$U+K = 5.0~J$
$K = 5.0~J - U$
$K = 5.0~J - 2.0~J$
$K = 3.0~J$
We can find the speed of the particle at point B;
$K = 3.0~J$
$\frac{1}{2}mv^2 = 3.0~J$
$v^2 = \frac{(2)(3.0~J)}{m}$
$v = \sqrt{\frac{(2)(3.0~J)}{m}}$
$v = \sqrt{\frac{(2)(3.0~J)}{0.50~kg}}$
$v = 3.5~m/s$
We can find the kinetic energy at point C;
$U+K = E$
$U+K = 5.0~J$
$K = 5.0~J - U$
$K = 5.0~J - 3.0~J$
$K = 2.0~J$
We can find the speed of the particle at point C;
$K = 2.0~J$
$\frac{1}{2}mv^2 = 2.0~J$
$v^2 = \frac{(2)(2.0~J)}{m}$
$v = \sqrt{\frac{(2)(2.0~J)}{m}}$
$v = \sqrt{\frac{(2)(2.0~J)}{0.50~kg}}$
$v = 2.8~m/s$
We can find the kinetic energy at point D;
$U+K = E$
$U+K = 5.0~J$
$K = 5.0~J - U$
$K = 5.0~J - 0$
$K = 5.0~J$
We can find the speed of the particle at point D;
$K = 5.0~J$
$\frac{1}{2}mv^2 = 5.0~J$
$v^2 = \frac{(2)(5.0~J)}{m}$
$v = \sqrt{\frac{(2)(5.0~J)}{m}}$
$v = \sqrt{\frac{(2)(5.0~J)}{0.50~kg}}$
$v = 4.5~m/s$
