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## 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 modelled 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?