A roller coaster glides from the rest from the top of an 80.0 m hill. What is the speed of the roller coaster at the bottom of the hill?

Respuesta :

As energy is conserve:

[tex]Epi+Ekl=Epf+Ekf[/tex]

i - initial

f - final

Beginning object at rest, [tex]Eki=0[/tex]. Object on the lowest position, hence [tex]Epf=0[/tex].

Therefore:

[tex]Epi=Epf[/tex]

[tex]M.g.h.=\frac{1}{2} (v^{2} )[/tex]

Assume no Mass changed:

[tex]g.h.=\frac{1}{2}( v^{2} )[/tex]

V is at bottom and assuming that [tex]g=\frac{10m}{s^2}[/tex]:

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2(10)(80)}[/tex]

[tex]v=\sqrt{1,600}[/tex]

[tex]v=40[/tex]

The velocity is [tex]40m/s[/tex]