Answer to Question #98462 in Classical Mechanics for Rachel

Question

Hi! Was wondering how you would solve this problem...here is the link

https://www.albert.io/learn/ap-physics-1-and-2/unit-4-energy/energy-of-descending-objects/kinetic-energy-on-mars

Let me know! Thanks :)

Answer 1

total mechanical energy at point A is equal to

$E_A=m\cdot g \cdot h$

total mechanical energy at point  $A^{'}$ is equal to

$E_{A^{'}}=m\cdot g \cdot x+K_{A^{'}}=m\cdot g \cdot x+0.6\cdot m\cdot g \cdot h$

full mechanical energy is stored, then we write

$E_A=E_{A^{'}}$

$m\cdot g \cdot h=m\cdot g \cdot x+0.6\cdot m\cdot g \cdot h$

$h= x+0.6\cdot h$

where from

$x= h-0.6\cdot h=2-0.6\cdot 2=1.2=0.8(m)$

from the likeness of triangles

$OAB end OA^{'}B^{'}$ we write

$\frac{AB}{A^{'}B^{'} }=\frac{OA}{OB^{'} }$

or

$\frac{h}{x}=\frac{5}{L }$

where from

$L=5\cdot\frac{x}{h}=5\cdot\frac{0.8}{2}=2(m)$

The desired point is located $x=0.8(m)$ , $L=2(m)$ .

Answer:

At 2 mfrom the bottom of the ramp.