nonconservative forces and work..just need someone to check if correct?

Question

The skeleton track has 15 turns and drops 104 m in elevation from top to bottom.

(a) in the absense of nonconservative forces such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed of the rider at the beginning of the run is relatively small and can be ignored.

so for this part you just solve for the final velocity which is:
vf=sqrt vo+2g(ho-hf)
since the initial velocity can be ignored..we dont use it, therefore:
sqrt 2(9.80)(104)=45.15 m/s?

(b) In reality, the best riders reach the bottom with a speed of 35.8 m/s. How much work is done on an 86-kg rider and skeleton by non-conservative forces?

so we use the equation:

Wnc=1/2m(vf^2-Vo^2)-mg(ho-hf)

=1/2(86.0kg)[(45.15)m/s^2-(35.8m/s)^2]-(86.0kg)(9.80 m/s^2)(104m)

=1/2(86.0kg)(756.88m/s)-87651.2

= -57376J

does it make sense to have a negative work in this case?

thanks!!

Answer 1

It does make sense to have negative work, but your answer for part (b) is wrong. You should have Wnc = 1/2(86.0)(35.8^2 - 0^2) - 86.0(9.80)(104) = -3.25×10^4 J.

However, given part(a) it's simpler to work out how much energy is missing: actual KE - expected KE = 1/2 (86.0) (35.8^2 - 45.15^2) = -3.25×10^4 J.

Now, the reason why we have negative work done: actually this should be pretty obvious if you think back to work = force times distance. The force is operating in one direction, but the displacement is in the opposite direction. So the work done on the rider is negative - which is why the rider's KE is decreased rather than increased.