- #1

Argonaut

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- Homework Statement
- [Sears and Zemansky's University Physics, 13E] 6.19/(e)

Use the work–energy theorem to solve the problem. Neglect air resistance.

At the base of a frictionless icy hill that rises at 25.0° above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

- Relevant Equations
- Work-Energy theorem

Kinetic Energy

Trigonometry

My solution is different from the official solution and I don't understand what I did wrong.

Here is my solution:

The magnitude of the initial velocity is ##|v_0| = 12.0~\rm{m/s}##, so the vertical component of the initial velocity is ##v_{0-y} = (12.0 \sin{25^{\circ}})~\rm{m/s}##.

Then I use the work-energy theorem and the work equation applicable in case of constant force and straight-line displacement.

##W = \vec{F} \cdot \vec{s}##

and

##W_{tot} = \Delta K = K_1 - K_0 = 0 - K_0 = -\frac{1}{2}v^2m##

Substituting the appropriate vertical variables, I obtain:

$$

-mgy=-\frac{1}{2}v_{0-y}^2m \\

$$

And then rearranging to make ##y## the subject of the equation I obtain

$$

y=\frac{v_{0-y}^2}{2g}=\frac{((12.0 \sin{25^{\circ}})~\rm{m/s})^2}{2(9.80~\rm{m/s^2})} = 1.31~\rm{m}

$$

However, the official solution says ##7.35~\rm{m}## which is the value I'd get if I just used the initial velocity magnitude, not its vertical component.

What am I missing?

[Edit: typo in homework statement]

Here is my solution:

The magnitude of the initial velocity is ##|v_0| = 12.0~\rm{m/s}##, so the vertical component of the initial velocity is ##v_{0-y} = (12.0 \sin{25^{\circ}})~\rm{m/s}##.

Then I use the work-energy theorem and the work equation applicable in case of constant force and straight-line displacement.

##W = \vec{F} \cdot \vec{s}##

and

##W_{tot} = \Delta K = K_1 - K_0 = 0 - K_0 = -\frac{1}{2}v^2m##

Substituting the appropriate vertical variables, I obtain:

$$

-mgy=-\frac{1}{2}v_{0-y}^2m \\

$$

And then rearranging to make ##y## the subject of the equation I obtain

$$

y=\frac{v_{0-y}^2}{2g}=\frac{((12.0 \sin{25^{\circ}})~\rm{m/s})^2}{2(9.80~\rm{m/s^2})} = 1.31~\rm{m}

$$

However, the official solution says ##7.35~\rm{m}## which is the value I'd get if I just used the initial velocity magnitude, not its vertical component.

What am I missing?

[Edit: typo in homework statement]

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