Q83P - University Physics with Modern Physics

Question

On an essentially frictionless, horizontal ice rink, a skater moving at encounters a rough patch that reduces her speed to due to a friction force that is 25% of her weight. Use the work–energy theorem to find the length of this rough patch

Step-by-Step Solution

Answer

The rough patch is 1.28 m long.

Step 1: Identification of given data

The given data can be listed below,

• The initial velocity of the skater is, ${\mathrm{v}}_{\mathrm{i}}=3\mathrm{m}/\mathrm{s}$ 
• The final velocity of the skater is, ${\mathrm{v}}_{\mathrm{f}}=1.65 \mathrm{m}/\mathrm{s}$
• The percentage of frictional force is,of weight. ${\mathrm{f}}_{\mathrm{k}}=25\% \mathrm{of} \mathrm{weight}$

Step 2: Concept/Significance of work done.

Energy transfer from one body to another is referred to as work. When a force moves an object, work is said to be done.

Step 3: Determination of the length of this rough patch

The length of the rough patch is obtained by the relation-

$$\begin{gathered} W_{tot}=-\mu mgl \\ -\mu mgl=\frac{1}{2}m\left(v_i^2-v_f^2\right) \\ l=\frac{1}{2\mu g}\left(v_i^2-v_f^2\right) \end{gathered}$$

Here, $\mu$ is the coefficient of friction, g is the acceleration due to gravity, $v_i$ is the initial velocity of the skater and is the final velocity of the skater.

For the given values, the above equation becomes-

$$\begin{gathered} \mathrm{l}=\frac{1}{2\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.25\right)}\left(3^2-{\left(1.65\right)}^2\right){\left(\mathrm{m}/\mathrm{s}\right)}^2 \\ =1.28\mathrm{m} \end{gathered}$$ 

Thus, the length of the rough patch is 1.28 m.