Total energy, speed, resistive force

[furor celtica]

Homework Statement

A skier of mass 70 kg sets off, with initial speed of 5 m(s^-1), down the line of greatest slope of an artificial ski-slope. The ski-slope is 80 metres long and is inclined at a constant angle of 20° to the horizontal. During the motion the skier is to be modeled as a particle.
- Ignoring air resistance and friction, calculate the speed of the skier at the bottom of the slope.
- The skier actually reaches the bottom of the slope with speed 6 m(s^-1). Calculate the magnitude of the constant resistive force along the slope which could account for this final speed.

Homework Equations

The Attempt at a Solution

For the first question my work looks like this:
(Total energy at the top of the slope) = (total energy at the bottom of the slope)
=> (PE at the top of the slope + KE at the top of the slope) = (PE at the bottom of the slope + KE at the bottom of the slope)
=> ((70g x 80sin20) + (0.5 x 70 x 5^2)) = (0 + (0.5 x 70 x v^2))
Taking g=10
=> v = sqrt(((56000sin20) + (35 x 25))/35) = 23.9 m(s^-1) (to 3 s.f.)

However, the correct answer is 23.4 m(s^-1)

For the second question my work looks like this:
(PE + KE) at the top of the slope - (80F) = (PE + KE) at the bottom of the slope
Where F is the constant resistive force
This gives (56000sin20) + (35 x 25) – 80F = 35 x 36
=> 80F = (56000sin20) + (35 x 25) – (35 x 36)
=> F = ((56000sin20) + (35 x 25) – (35 x 36))/80
=> F = 234.6 N (to 1 d.p.)

However, the correct answer is 224 N Newtons (to 3 s.f. probably).

What am I missing here?

 

[ehild]

Use g=9.8 m/s^2 instead of 10. ehild

 

[furor celtica]

i know I'm supposed to use g=10 but i used g=9.8 anyway and that gives 23.7 ms^-1 and 230 N, which are both incorrect.
anything else?

 

[ehild]

I think your results are the correct ones. I have got the same.

ehild

 

[furor celtica]

then i wasted literally hours making the same calculations over and over. yay for textbook typos.
thanks for your help