Chapter 6, Problem 48P

Part (a)

To determine

Speed of the ski.

Answer to Problem 48P

Solution: 25.5 m/s

Explanation of Solution

Given:

Initial speed of ski = $\text{0}\text{m/s}$

Length of inclined = $\text{85}\text{m}$

Angle of inclination = $28°$

Coefficient of friction = 0.090

Formula used:

Apply conservation of energy:

  $W_{NC}=\Delta KE+\Delta PE$

Work done by non-conservative force: $W_{NC}$

Change in kinetic energy, $\Delta KE$

Change in potential energy, $\Delta PE$

Calculation:

  

Apply Newton’s first law perpendicular to inclined plane:

  $\begin{array}{c}N=mg\cos {28}^0\\ W_{NC}=\overrightarrow{F\cdot }\overrightarrow{S}\\ =FS\cos 180°\\ =-FS\\ =-\left(\mu N\right)\left(S\right)\\ =-0.090\times mg\cos 28°\times 85\end{array}$

  $\begin{array}{c}{W}_{NC}=-0.090mg\cos 28°\times 85----\left(1\right)\\ \left(P{E}_1\right)=mg\times 85\cos {28}^0\\ \text{K}\text{.E}{\text{.}}_1=\frac{1}{2}m{v}_1{}^2\\ =\frac{1}{2}m{\left(0\right)}^2\\ =0\text{J}\end{array}$

Final potential energy

  $\begin{array}{c}\left(P{E}_2\right)=mg\left(0\right)\\ =\text{0J}\end{array}$

  $\begin{array}{c}\left(K{E}_2\right)=\frac{1}{2}m{v}_2{}^2\\ W_{NC}=\Delta PE+\Delta KE\\ =(K{E}_2-K{E}_1)+(P{E}_2-P{E}_1)\end{array}$

  $\begin{array}{c}{v}_2=\sqrt{2\times 9.80\times 85\left(\sin 28°-0.090\cos 28°\right)}\\ \text{=25}\text{.5}\text{m/s}\end{array}$

Conclusion:The speed of the ski is 25.5 m/s.

Part (b)

To determine

The distance that the ski travel along the level.

Answer to Problem 48P

Solution:368 m

Explanation of Solution

Given:

Initial speed of ski = $\text{0}\text{m/s}$

Length of inclined = $\text{85}\text{m}$

Angle of inclination = $28°$

Coefficient of friction = 0.090

Formula used:

Apply conservation of energy

  $W_{NC}=\Delta KE+\Delta PE$

Work done by non-conservative force: $W_{NC}$

Change in kinetic energy, $\Delta KE$

Change in potential energy, $\Delta PE$

Calculation:

Again, apply conservation of energy

  $\begin{array}{l}{W}_{NC}=\Delta KE+\Delta PE\\ W_{NC}=-\mu N\times d\end{array}$

Where, d is the horizontal distance travelled.

  $\begin{array}{c}\Delta KE=\frac{1}{2}m{v}_2{}^2-\frac{1}{2}m{v}_1{}^2\\ =\frac{1}{2}m{\left(0\right)}^2-\frac{1}{2}m{\left(\text{25}\text{.5}\right)}^{\text{2}}\\ \text{=-}\frac{1}{2}m{\left(\text{25}\text{.5}\right)}^2\end{array}$

Because level of block does not change, $\Delta PE=0$

Substitute the above value in conservation of energy equation

  $\begin{array}{l}-\left(0.090mg\right)\times d=-\frac{1}{2}m{\left(25.5\right)}^2\\ d=368\text{m}\end{array}$

Conclusion:So, distance travelled = 368 m.