A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m long. If the force of friction as she rides down the hill is 20N, what us the speed at the bottom.

13.78 m/s

Explanation:

Given that:

The mass of the girl & her bicycle = 42 kg

The speed at the top of the hilll= 4 m/s

The height of the hill = 14.3 m

The length of the hill (distance) = 112 m

The frictional force = 20 N

To find the speed at the bottom of the hill; we need to carry out the following processes.

The workdone by gravity = mass × acceleration due to  gravity × Δh

= 42 × 9.8 × ( 14.3 - 0 )

= 5885.88 joules

The workdone by the friction = Force × distance = - 20 × 112    (since she is riding down the hill)

The workdone by the friction = -2240 Joules

The initial Kinetic friction = 1/2 mv²

= 1/2 × 42 × 4²

= 336 Joules

The final kinetic energy = Initial Kinetic energy + total work  

The final kinetic energy = (336 + 5885.88 - 2240) Joules

The final kinetic energy = 3981.88   Joules

Using the final kinetic energy =   1/2 mv²

3981.88  = 1/2 × 42 × v²

3981.88  = 21 v²

v² = 3981.88/21

v² =189.61

v =

v = 13.78 m/s

Therefore, the speed at the bottom = 13.78 m/s

i already answered it by 34/s