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Exercises

Energy conservation

Exercise 5.3

A \(\text{60,0}\) \(\text{kg}\) skier with an initial speed of \(\text{12,0}\) \(\text{m·s$^{-1}$}\) coasts up a \(\text{2,50}\) \(\text{m}\)-high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is \(\text{0,0800}\). (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

We need to determine the length of the slope as this is the distance over which friction acts as well as the normal force of the skier on the slope to determine the magnitude of the force due to friction. The normal force balances the component of gravity perpendicular to the slope, therefore: \begin{align*} F_{\text{friction}}&= - \mu N \\ &= - \mu F_g\cos\theta \\ &= - \mu mg \cos\theta \end{align*} The length of the slope will be \(\Delta x = \dfrac{h}{\sin\theta}\).

\begin{align*} W_{\text{non-conservative}} + E_{k,i} + E_{p,i} & = E_{k,f} + E_{p,f} \\ -\mu mg\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}mv_i^2 +mgh_i &= \frac{1}{2}mv_f^2 + mgh_f \\ -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i &= \frac{1}{2}v_f^2 + gh_f \\ \frac{1}{2}v_f^2 + gh_f &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i \\ \frac{1}{2}v_f^2 + &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +g(h_i -h_f) \\ v_f^2 &= -2\mu g\cos\theta \dfrac{h}{\sin\theta} + v_i^2 +2g(h_i -h_f) \end{align*} \begin{align*} v_f^2 &= -2(0,08)(9,8) \cos(35) \dfrac{2.5}{\sin(35)} + (12)^2 +2(9,8)(-2,5) \\ v_f & = \sqrt{89,4016598} \\ v_f & = \text{9,46}\text{ m·s$^{-1}$} \end{align*} \(\text{9,46}\) \(\text{m·s$^{-1}$}\)
How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is \(\text{110}\) \(\text{km·h$^{-1}$}\)?

\begin{align*} 110 km/hr&= 110 \times \frac{1000}{3600} \\ &= \text{30,56}\text{ m·s$^{-1}$} \end{align*} \begin{align*} mgh&= \frac{1}{2}mv^2 \\ h&= \frac{v^2}{2g} \\ &= \frac{(\text{30,56})^2}{2(\text{9,8})} \\ &= \text{47,65}\text{ m} \end{align*}

\(\text{47,65}\) \(\text{m}\)
If, in actuality, a \(\text{750}\) \(\text{kg}\) car with an initial speed of \(\text{110}\) \(\text{km·h$^{-1}$}\) is observed to coast up a hill to a height \(\text{22,0}\) \(\text{m}\) above its starting point, how much thermal energy was generated by friction?

Kinetic energy was converted into potential energy. The addition of friction as a dissipative force ensures that some of the kinetic energy is lost as thermal energy. The difference in the potential energy gained with and without friction is the energy lost to friction. The gravitational potential energy gained is only related to height so the difference in height allows us to determine the energy lost to friction quickly.

Without friction the car rose to a height of \(\text{47,65}\)\(\text{m}\), with friction the height was only \(\text{22}\)\(\text{m}\). The energy lost to friction is equivalent to the energy required to increase the gravitational potential energy by raising the vehicle a height of \(\text{25,65}\)\(\text{m}\). Therefore the energy lost to friction is:

\begin{align*} W&= mgh \\ &= (750)(9,8)(25.65) \\ & = \text{188 527,5}\text{ J} \end{align*}

We calculated the gravitational potential energy that was lost as friction but the correct answer will be negative this quantity as the work done by friction is negative: -\(\text{188 527,5}\) \(\text{J}\)

What is the average force of friction if the hill has a slope \(\text{2,5}\) \(\text{º}\) above the horizontal?

We will assume a constant force of friction. The disance over which the friction acted is the length of the slope, we know the angle of the slope and the vertical height so we can calculate the distance (hypotenuse). We can use the definition of work to calculate the force, remember that friction acts opposite to the displacement:

\begin{align*} W_{friction}&= F_{friction}\Delta x \cos\theta \\ &= -F \Delta x \\ &= -F \frac{h}{\sin(2.5)} \\ -\text{188 527,5}&= -F \frac{h}{\sin(2.5)} \\ F&= \text{188 527,5} \frac{\sin(2.5)}{22} \\ F&= \text{373,79}\text{ N} \end{align*} \(\text{373,79}\) \(\text{N}\)
A bullet traveling at 100 m/s just pierces a wooden plank of 5 m. What should be the speed (in m/s) of the bullet to pierce a wooden plank of same material, but having a thickness of 10m?

Final speed and hence final kinetic energy are zero in both cases. From "work- kinetic energy" theorem, initial kinetic energy is equal to work done by the force resisting the motion of bullet. As the material is same, the resisting force is same in either case. If subscript "1" and "2" denote the two cases respectively, then:

5m:

\begin{align*} 0−\frac{1}{2}mv_1^2 & = −Fx_1 \\ \frac{1}{2}m(100)^2=F\times 5 \end{align*}

10m:

\begin{align*} 0−\frac{1}{2}mv_2^2 & = −Fx_2 \\ \frac{1}{2}mv_2^2=F\times 10 \end{align*}

Take the ratio of these two equations:

\begin{align*} \frac{v_2^2}{(100)^2} &=\frac{F\times 10}{F\times 5}=2 \\ v_2^2 & = 2 \times 10000 = 20000 \\ v_2 & = \text{141,4}\text{ m·s$^{-1}$} \end{align*}

\(\text{141,4}\) \(\text{m·s$^{-1}$}\)