Question

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?

Answer 1

Answer:

The bikers speed at the top of other hill is 25.82 m/s.

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

$h_{1}$ = 44m

$h_{2}$ = 10m

Since the biker starts from rest , his initial speed $v_{1}$ = 0 m/s

Let final speed of the bike at the top of other hill be $v_{2}$.

∴ Initial Energy (at the top of 44m hill) = $mgh_{1}$

  Final Energy  (at the top of 10m hill) =  $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$.

Conserving both the energies , we get

$mgh_{1}$ = $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$

∴ $v_{2} = \sqrt{2g(h_{1}-h_{2} )}$

Substituting the values for g , $h_{1}$ , $h_{2}$ , we get

$v_{2}$ = 25.82 m/s