Question

A 60 kg skier with an initial speed of 16 m/s Coasts up a 2.5 m high rise as shown. Find her final speed at the top in meters per second, given that the ice ground is frictionless

Answer 1

This problem relies on the law of conservation of energy

The equation below is the formula for kinetic energy

$KE=\frac{1}{2}mv^2$

And the formula for gravitational potential energy is

$PE=mgh$

So to find final speed, we will use both of these formulas

$\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2$$\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2$

This formula gives us the energy before and after the skier moves up the hill.

For variables, we have

v1 = 16 m/s

mass = 60 Kg

h2 = 2.5 m

h1 = 0 m

$\begin{gathered} \frac{1}{2}(60)(16)^2+(60)g(0)=\frac{1}{2}(60)(v_2)^2+(60)g(2.5) \\ 256=v_2^2+49 \\ v_2^2=207 \\ v_2=14.38\text{ }\frac{m}{s} \end{gathered}$

Final Velocity is 14.38 m/s