Calculate the deceleration of a snow boarder going up a 5.0 slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 5.9 may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies.

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.)

explanation:

areas w steep slopes, west coast, rocky mountains, and appalachian mountains

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.