A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so much fun coasting that he decided not to pedal even climbing the next hill. If that hill is 10 m high, what will the biker s speed be at the top of that hill?

The problem can be solve by applying the law of conservation of energy. In absence of frictional forces, the total mechanical energy of the bike (the sum of potential energy + kinetic energy) must be conserved. So we can write:

[tex]U_i +K_i = U_f + K_f[/tex]

where

[tex]U_i[/tex] is the initial potential energy at the top of the first hill

[tex]K_i[/tex] is the initial kinetic energy at the top of the first hill

[tex]U_f[/tex] is the final potential energy at the top of the second hill

[tex]K_f[/tex] is the final kinetic energy at the top of the second hill

We can rewrite the equation as:

[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]

where:

m is the mass of the bike

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

[tex]h_i = 44 m[/tex] is the height of the first hill

u = 0 m/s is the speed at the top of the first hill

[tex]h_f = 10 m[/tex] is the height of the second hill

v is the speed at the top of the second hill

And solving for v, we find:

[tex]mgh_i = mgh_f + \frac{1}{2}mv^2\\v^2=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(44-10)}=25.8 m/s[/tex]

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