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