Question

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed by 50 % due to a friction force that is 22 % of her weight. Use the work–energy theorem to find the length of this rough patch

Answer 1

The length of the rough patch is 4.345 meters.

Step-by-step explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

`K_(1) = K_(2) + W_(loss)`

Where:

`K_(1)`,
`K_(2)` - Initial and final kinetic energy, measured in joules.

`W_(loss)` - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

`(1)/(2)\cdot m \cdot v_(1)^(2) = (1)/(2)\cdot m \cdot v_(2)^(2) + f \cdot \Delta s`

Where:

`v_(1)`,
`v_(2)` - Initial and final speed of the skater, measured in meters per second.

`m` - Mass of the skater, measured in kilograms.

`\Delta s` - Length of the rough patch, measured in meters.

`f` - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

`f = r \cdot m \cdot g`

Where:

`r` - Ratio of friction force to weight, dimensionless.

`g` - Gravitational constant, measured in meters per square second.

Then,

`(1)/(2)\cdot m \cdot v_(1)^(2) = (1)/(2)\cdot m \cdot v_(2)^(2) + r \cdot m \cdot g \cdot \Delta s`

The equation is simplified algebraically and patch length is cleared afterwards:

`(1)/(2)\cdot (v_(1)^(2)-v_(2)^(2)) = r \cdot g \cdot \Delta s`

`\Delta s = (v_(1)^(2)-v_(2)^(2))/(2 \cdot r \cdot g )`

Given that
`v_(1) = 5\,(m)/(s)`,
`v_(2) = 2.5\,(m)/(s)`,
`r = 0.22` and
`g = 9.807 \,(m)/(s^(2))`, the length of the rough patch is:

`\Delta s = (\left(5\,(m)/(s) \right)^(2)-\left(2.5\,(m)/(s) \right)^(2))/(2\cdot (0.22)\cdot \left(9.807\,(m)/(s^(2)) \right))`

`\Delta s = 4.345\,m`

The length of the rough patch is 4.345 meters.