Q22E - University Physics with Modern Physics

Question

Use the work-energy theorem to solve each of these problem. You can use Newton’s laws to check your answers. (a) A skier is moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping? (b) Suppose the rough patch in part (a) was only 2.90 m long. How fats would the skier be moving when she reached the end of the patch? (c) At the base of a frictionless icy hill that rises at 25.0⁰ 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?  

Step-by-Step Solution

Answer

(a)  5.8 m

(b)  3.53ms^-1

(c)  17.38 m

Step 1: Identification of the given data

The given data is listed below as-

• The initial velocity of skier 5 is $m{s}^{-1}$ 
• Coefficient of kinetic friction is ${\mu }_k=0.220$ 
• The length of rough patch is 2.90 m
• The speed of toboggan is 12.0 m/s

Step 2: Significance of the work energy theorem

When forces act on a particle it undergoes displacement and the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. Therefore, work done on the particle is given by-

${\mathbf{W}}_{\mathbf{totaL}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{‐}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{∆}\mathbf{K}$ 

The work-energy theorem can be applied to all the bodies that can be treated as particles.

Step 3: Determination of distance travelled by skier on the patch

(a)

Use Newton’s law to calculate the distance by the skier on the patch 

Now, if initial, final velocity and acceleration are given then we can use the following equation:

${v^2}_f={v^2}_i+2as$  ………….(1)

Here, $v_i=v_{1A}$ and $v_f=v_{2B}$ are the initial and final velocity of the particle respectively.

Therefore, equation (1) will become

${v^2}_{2B}={v^2}_{1A}+2as$ 

$v_{2B}=0$ since we are calculating the distance travelled by skier on this patch.

Therefore,  $0={v^2}_{1A}+2as$

Rearrange the above equation to get,

$s=-\frac{{V^2}_{1A}}{2a}$ ………..(2) 

To calculate a in the direction of motion we have:

$F=-f_k$ 

Now, force due to motion of the skier in +x direction is $F=ma$ and force due to kinetic friction in -x direction is $F=-f_k$  

Therefore, $ma=-{\mu }_{k}mg$  

Or, $a=-{\mu }_{k}g$  

So, equation (2) will now become

$s=\frac{{v_{1A}}^2}{2{\mu }_{k}g}$  

The final velocity of the skier is:

$v_{2B}=\sqrt{{v_1}^2 a+2as}$ 

Substitute the value of acceleration $a=-{\mu }_k g$ in above equation

$v_{2B}=\sqrt{{v_1}^2 a-2{\mu }_{k}gs}$ 

The skier is moving at $5 m{s}^{-1}$ and it encounters a long, rough horizontal patch of snow which has coefficient of kinetic friction as ${\mu }_k=0.220$ .

Therefore, the initial speed of the skier is $5 m{s}^{-1}$ 

And final speed of the skier is $0 m{s}^{-1}$ 

It is required to determine the length of rough patch.

Now, from equation (2), length of patch can be obtained.

$$\begin{gathered} s=-\frac{{V^2}_{1A}}{2a} \\ s=-\frac{{\left(5 {\mathrm{ms}}^{-1}\right)}^2}{2\times 0.22\times 9.8 {\mathrm{ms}}^{-2}} \\ =5.8 \mathrm{m} \end{gathered}$$   

    

Thus, the distance travelled by skier on the patch is 5.8 m 

Step 4: Determination of speed of the skier when reached at the end of patch

(b)

It is given that length of rough horizontal patch of snow is $s=2.9 \mathrm{m}$ and has a coefficient of friction as ${\mu }_k=0.220$ 

Therefore, the initial speed of the skier is $5 m{s}^{-1}$ 

And it is required to find the final velocity  $v_{2B}$ of the skier 

From equation (2)

$$\begin{gathered} s=-\frac{{V^2}_{1A}}{2a} \\ {v}_{2B}=\sqrt{{\left(5 {\mathrm{ms}}^{-1}\right)}^2-2\times 0.22\times 9.8 {\mathrm{ms}}^{-2}\times 2.9 \mathrm{m}} \\ =3.53{\mathrm{ms}}^{-1} \end{gathered}$$ 

  

Thus, the speed of the skier when she reached at the end of the patch is $3.53{\mathrm{ms}}^{-1}$  

Step 5: Determination of distance travelled by toboggan above the base before stopping

(c)

It is given that the base of a frictionless icy hill rises at 25.0⁰ above the horizontal and a toboggan has a speed of 12 ms^-1 toward the hill.

Therefore, the initial velocity of the toboggan is $v_{1A}=12 {\mathrm{ms}}^{-1}$ and final velocity $v_{2B}=0 {\mathrm{ms}}^{-1}$ 

The total force in this case is:

$$\begin{gathered} F_{tot}=ma \\ =mg \sin \theta \end{gathered}$$ 

Take the x-axis along inclined plane, therefore the angle $\varphi =180°$ .

So, the work done opposite to the velocity will be

  $$\begin{gathered} W_{tot}={\overrightarrow{F}}_{tot}·\overrightarrow{s} \\ =F_{tot} s \cos \varphi \\ =F_{tot} s \cos 180° \\ =-mgs \sin \theta \end{gathered}$$

Now, change kinetic energy is given by

$W=K_2-K_1$  

Therefore,

$-mg \sin \theta =\frac{1}{2}m{\left(0\right)}^2-\frac{1}{2}m{\left(v_{1A}\right)}^2$  

Rearrange the above equation:

$$\begin{gathered} s=\frac{{v^2}_{1A}}{2g \sin \theta } \\ =\frac{{\left(12 {\mathrm{ms}}^{-1}\right)}^2}{2\times 9.8 {\mathrm{ms}}^{-2}\times \sin 25°} \\ =17.38 \mathrm{m} \end{gathered}$$ 

Thus, distance travelled by toboggan above the base before stopping is $17.38 \mathrm{m}$.