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Q5.1 10PE

Expert-verifiedFound in: Page 188

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Calculate the deceleration of a snow boarder going up a ${\mathbf{\text{5.0}}}$ ****, slope assuming the coefficient of friction for waxed wood on wet snow. The result of ****Exercise ${\mathbf{\text{5.1}}}$**** ****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**.** **

Deceleration of snow boarder is ${\text{1.83 m/s}}^{\text{2}}$ . It is also represented by ${\text{a = - 1.83 m/s}}^{\text{2}}$ because its direction is opposite that of motion.

The negative sign implies that it is moving in the opposite direction.

**When acceleration and velocity point in opposing directions, this is called deceleration. Deceleration, like acceleration, can be expressed in both positive and negative terms.**

Friction opposes relative motion between the surfaces so if snow boarder going up then friction will oppose this upward motion.

Snow boarder is going up so there is relative motion between the surface of snow boarder and waxed wood. Friction force opposes the relative motion. Therefore, in this case direction of kinetic friction will be down the incline.

As block is at rest perpendicular to the inclined surface, therefore net force perpendicular to the inclined

$\mathrm{N}=\mathrm{mgcos}\left(\mathrm{\theta}\right)$

So kinetic friction, $\mathrm{f}={\mathrm{\mu}}_{\mathrm{K}}\mathrm{N}={\mathrm{\mu}}_{\mathrm{K}}\mathrm{mgcos}\left(\mathrm{\theta}\right)$

Component of gravitational force down the incline $=\mathrm{mgsin}\left(\mathrm{\theta}\right)$

Net downward force along the inclined

$\mathrm{F}=\mathrm{mgsin}\left(\mathrm{\theta}\right)+{\mathrm{\mu}}_{\mathrm{K}}\mathrm{mgcos}\left(\mathrm{\theta}\right)$

As it is in downward direction and snow boarder is moving up so due to this net there will be deceleration which can be calculated by using Newton’s second law

$\mathrm{ma}=\mathrm{mgsin}\left(\mathrm{\theta}\right)+{\mathrm{\mu}}_{\mathrm{K}}\mathrm{mgcos}\left(\mathrm{\theta}\right)$

or $\mathrm{a}=\mathrm{gsin}\left(\mathrm{\theta}\right)+{\mathrm{\mu}}_{\mathrm{K}}\mathrm{gcos}\left(\mathrm{\theta}\right)$ (from table role="math" localid="1668662577574" $5.1{\mathrm{\mu}}_{\mathrm{K}}$ waxed wood on wet snow $=0.1$)

or role="math" localid="1668662705741" $\mathrm{a}=9.8\mathrm{sin}\left({5}^{\circ}\right)+0.1\times 9.8\times \mathrm{cos}\left({5}^{\circ}\right)$

or ${\text{a = 1.83 m/s}}^{\text{2}}$

So, deceleration of snowboarder is ${\text{1.83 m/s}}^{\text{2}}$ . It is also represented by because its direction is opposite that of motion.

The negative sign implies that it is moving in the opposite direction.

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