Let θ be the angle of the slope (looks like θ = 5.0°?) and µ the coefficient of kinetic friction between waxed wood and wet snow (not given).
As the snowboarder ascends the slope, there are 3 forces acting on them:
• their weight, pulling them downward
• the normal force of the slope pushing up on the snowboarder, perpendicular to the slope
• friction that opposes the up-slope motion, parallel to the slope
Split up the forces into components acting parallel or perpendicular to the slope:
• net parallel force:
∑ F = - f - mg sin(θ) = - ma
(we take the up-slope direction to be positive, so both friction f and the parallel component of the snowboarder's weight points down-slope. We expect the snowboarder to slow down as they move up the slope, so acceleration points in the same direction as friction)
• net perpendicular force:
∑ F = n - mg cos(θ) = 0
We have
f = µ n
so that
n = mg cos(θ) → f = µmg cos(θ)
→ - µmg cos(θ) - mg sin(θ) = - ma
→ - µg cos(θ) - g sin(θ) = - a
→ a = (µ cos(θ) + sin(θ)) g
Now just plug in the parameters you probably know from Exercise 5.9. (Notice that a is positive here since it is a magnitude, but it points down the hill.)
