A roller coaster has a mass of 500 kg. It drops from rest at the top of a hill that's 57 m tall. How fast is it going when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2.

Respuesta :

The answer should be 33.4m/s

Answer:

[tex]v=33.42 \frac{m}{5}[/tex]

Explanation:

Hello

The principle of conservation of energy indicates that energy is not created or destroyed; It only is  transformed in some ways into others. In these transformations, the total energy remains constant; that is, the total energy is the same before and after each transformation.in this case the  potential energy is the mechanical energy associated with the location(heigth) Ec=mgh, here m is the mass, g is the acceleration due to gravity and h is the height

[tex]E_{g}=mgh[/tex]

and , it must be equal to the kinetic energy in the bottom, it is given by

[tex]E_{k}=\frac{mv^{2}}{2}\\[/tex]

where m is the mass of the object , and v its velocity

According to the principle of conservation of energy

[tex]E_{g}=E_{k}\\mgh=\frac{mv^{2} }{2} \\Now, isolate\ v\\gh=\frac{v^{2} }{2} \\2gh=v^{2}\\\sqrt{2gh} =v\\[/tex]

the velocity only can be positive, it means we need just the positive root

Replacing

[tex]v= \sqrt{2gh} \\v= \sqrt{2(9.8 \frac{m}{s^{2} } )*57m} \\v=\sqrt{\frac{117.2m^{2} }{s^{2} } } \\v=33.42 \frac{m}{5}[/tex]

Have a great day

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