A 60-kg skier pushes off the top of a frictionless hill with an initial speed of 4.0 m/s.How fast will she be moving after dropping 10 m in elevation? Air resistance is negligible.
A)0.20 km/s
B)15 m/s
C)10 m/s
D)0.15 km/s
E)49 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we use the principle of conservation of mechanical energy, which states that the total mechanical energy (kinetic energy + potential energy) in a system remains constant if only conservative forces are doing work. The skier's initial kinetic energy plus her initial potential energy equals her final kinetic energy (since her final potential energy at the bottom is zero in this context). The equation for kinetic energy is $KE = \frac{1}{2}mv^2$ and for potential energy is $PE = mgh$, where $m$ is mass, $v$ is velocity, $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$), and $h$ is height.Initial kinetic energy: $KE_i = \frac{1}{2} \times 60 \times (4)^2 = 480 J$Initial potential energy: $PE_i = 60 \times 9.8 \times 10 = 5880 J$Total initial energy: $E_i = KE_i + PE_i = 480 J + 5880 J = 6360 J$At the bottom, all this energy is converted into kinetic energy:$6360 J = \frac{1}{2} \times 60 \times v^2$Solving for $v$, we find:$v^2 = \frac{6360 \times 2}{60}$$v^2 = 212$$v = \sqrt{212} \approx 14.56 m/s$Rounding to the nearest whole number gives us approximately 15 m/s.

A 60-Kg Skier Pushes Off the Top of a Frictionless