A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5.0 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N, what is her speed at the bottom

v =   11 m/s   is her final speed

Explanation:

work done by gravity = m g Δh =   40×9.8×10   = 3920 Joules

Work done by friction = - force×distance =   - 20×100   =   - 2000 Joules

[minus sign because friction force is opposite to the direction of motion]

Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2   = 500 Joules

Now, by work energy theorem

Work done = change in kinetic energy.

Final K.E. = initial K.E. + total work =    500 + 3920 - 2000  = 2420 Joules

Now, Final K.E. = (1/2) m v^2  [final speed being v= speed at the bottom]

⇒  2420 = (1/2)×40×v^2

   ⇒  121 = v^ 2

  v =   11 m/s   is her final speed

the answer is b ) square root of 5