SOLUTION

$($a$)$ Find the distance the block travels up the ramp.

Apply conservation of mechanical energy.

Substitute $v_i=v_f=0,y_i=0,y_f=h=$ dsin$,$ and $x_f=0.$

Solve for the distance $d$ and insert the known values.

$\frac{1}{2}m{v}_i^2+mg{y}_i+\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+mg{y}_f+\frac{1}{2}k{x}_f^2$

$\frac{1}{2}k{x}_i^2=mgh=$mgdsin$$

$d=\frac{\frac{1}{2}k{x}_1^2}{mgsin}=\frac{\frac{1}{2}(625\frac{N}{m})(-0\mathrm{.}100(m){)}^2}{(0\mathrm{.}500(kg))(9\mathrm{.}80\frac{m}{s^2})sin(30\mathrm{.}0)}$

$=1\mathrm{.}28m$

$($b$)$ Find the velocity at half the height, $\frac{h}{2}.$ Note that $h=$dsin$=(1\mathrm{.}28m)sin30\mathrm{.}0=0\mathrm{.}640m.$

Use energy conservation again.

$\frac{1}{2}m{v}_i^2+mg{y}_i+\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+mg{y}_f+\frac{1}{2}k{x}_f^2$

Take $v_i=0,y_i=0,v_f=\frac{1}{2}h,$ and $x_f=$

$\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+mg(\frac{1}{2}h)$

Multiply by $\frac{2}{m}$ and solve for $v_f.$

$(\frac{k}{m})x_i^2=v_f^2+gh$

$v_f=\sqrt[\phantom{2}]{(\frac{k}{m})x_i^2-gh}$

$=\sqrt[\phantom{2}]{(625\frac{N}{m}\frac{?}{0\mathrm{.}500kg})(-0\mathrm{.}100m{)}^2-(9\mathrm{.}80\frac{m}{s^2})(0\mathrm{.}640m)}$

$v_f=2\mathrm{.}50\frac{m}{s}$

Use the worked example above to help you solve this problem. A $0\mathrm{.}589kg$ block rests on a horizontal, frictionless surface as shown in the figure. The block is pressed back against a spring having a constant of $k=584\frac{N}{m},$ compressing the spring by $8\mathrm{.}5cm$ to point $($A$).$ Then the block is released.

$($a$)$ Find the maximum distance $d$ the block travels up the frictionless incline if $=30\mathrm{.}0.$

C

Your response differs from the correct answer by more than $10\%.$ Double check your calculations. m

$($b$)$ How fast is the block going when halfway to its maximum height?

$\frac{m}{s}$

EXERCISE

HINTS:

GETTING STARTED

I$\text{'}$M STUCKI

A $0\mathrm{.}88kg$ block is shot horizontally from a spring, as in the example above, and travels $0\mathrm{.}537m$ up a long a frictionless ramp before coming to rest and sliding back down. If the ramp makes an angle of $45\mathrm{.}0$ with respect to the horizontal, and the spring originally was compressed by $0\mathrm{.}1m,$ find the spring constant.

$\frac{N}{m}$

PRACTICE IT

Use the worked example above to help you solve this problem. A skier starts from rest at the top of a frictionless incline of height $20\mathrm{.}0m,$ as shown in the figure. As the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is $0\mathrm{.}217.$

$($a$)$ Find the skier's speed at the bottom.

$\frac{m}{s}$

$($b$)$ How far does the skier travel on the horizontal surface before coming to rest?

$m$

EXERCISE

HINTS: GETTING STARTED

I$\text{'}$M STUCK!

Use the values from PRACTICE IT to help you work this exercise. Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to $0\mathrm{.}217.($The incline makes an angle $=20\mathrm{.}0$ with the horizontal,$)$

Compute the distance traveled when there is no friction on the incline. When there is friction on the incline, do you expect the distance traveled to be larger or smaller than this? $m$

SOLUTION

$($a$)$ Let $y=0$ at $($B$).$ Calculate the potential energy at $($A$)$ and at $($B$),$ and calculate the change in potential energy.

Find $P{E}_i,$ the potential energy at $(4).$

$P{E}_i=mg{y}_i=(60\mathrm{.}0kg)(9\mathrm{.}80\frac{m}{s^2})(10\mathrm{.}0m)=$

$5\mathrm{.}88{10}^{3}J$

$P{E}_f=0$ at $(B)$ by choice. Find the

$P{E}_f-P{E}_i=0-5\mathrm{.}88{10}^{3}J=-5\mathrm{.}88{10}^{3}J$

difference in potential energy between

$($A$)$ and B$).$

$($b$)$ Repeat the problem if $y=0$ at $($A$),$ the new reference point, so that PE $=0$ at $($A$).$

Find $P{E}_f,$ noting that point $B$ is now at $y=-10\mathrm{.}0m.$

$P{E}_f=mg{y}_f=(60\mathrm{.}0kg)(9\mathrm{.}80\frac{m}{s^2})(-10\mathrm{.}0m)=$

$-5\mathrm{.}88{10}^{3}J$

$P{E}_f-P{E}_i=-5\mathrm{.}88{10}^{3}J-0=-5\mathrm{.}88{10}^{3}J$

$($c$)$ Repeat the problem, if $y=0$ two meters above $($B$).$

Find