Question

1. -/1 points a dog with a mass of 47.0 kg slides down a wet slope with negligible friction. the dog starts from rest and has a speed of 2.80 m/s at the bottom. what is the height of the slope (in m)?

Explanation:

Step1: Apply conservation of mechanical energy

Initial mechanical energy $E_1 = mgh$ (potential energy only as starts from rest, $v = 0$), final mechanical energy $E_2=\frac{1}{2}mv^{2}$ (potential energy is 0 at bottom, only kinetic energy). Since there is no friction, $E_1 = E_2$. So $mgh=\frac{1}{2}mv^{2}$.

Step2: Solve for height $h$

Cancel out mass $m$ from both sides of the equation $mgh=\frac{1}{2}mv^{2}$, we get $gh=\frac{1}{2}v^{2}$. Then $h = \frac{v^{2}}{2g}$. Given $v = 2.80\ m/s$ and $g = 9.8\ m/s^{2}$, substitute the values: $h=\frac{(2.80)^{2}}{2\times9.8}$.
$h=\frac{7.84}{19.6}=0.4\ m$

Answer:

$0.4$