# A girl rides a bicycle down a slope that is 2.55 m high and 12.6 m long (along the slope). Her speeds at the top and bottom of the slope are 1.35 m/s and 5.89 m/s respectively. How much work did...

A girl rides a bicycle down a slope that is 2.55 m high and 12.6 m long (along the slope). Her speeds at the top and bottom of the slope are 1.35 m/s and 5.89 m/s respectively. How much work did she do?

The total mass of the girl and bicycle is 43.0 kg. If air resistance and rolling resistance together are equivalent to a constant frictional force of 40.8 N, and her speeds at the top and bottom of the slope are 1.35 m/s and 5.89 m/s respectively, how much work did she do?

I understand to use the conservation of energy equation which would end up being

mgy + 1/2mvi^2 = 1/2mvf^2, and i know to use the frictional force for -fd, but i don't really understand where to go from there to get the work...if work is the change in kinetic energy, why can't i just take the velocities and find it that way?

### 1 Answer | Add Yours

This is related to the conservation of energy.

Let us assume the datum to be the horizontal line at the bottom of the slope.

When she is at top of the slope her energy is the kinematic energy of the bike and herself and the potential energy due to height.

When she comes down at the bottom her energy is the kinematic energy of the system.

Some energy will lost due o friction.

While she comes down she has to do some work against the friction.

So if you apply energy conservation here;

Initial energy+work she has done = Final energy+energy for friction

Initial energy = 1/2*43*1.35^2+43*9.81*2.55 = 1114.85 J

Final energy = 1/2*43*5.89^2 = 745.88 J

Friction energy = 40.8*12.6 = 514.08 J

Work she has done = 745.88+514.08-1114.85

= 145.11 J

So he has done a work of 145.11 J while riding the bicycle.

Note;

Usually we don't need to do work to ride down slope. Here since the friction force is very high she has to do some work. We can see that is fairly small compared to other energies.

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