Question

the miners working in a salt mine use smooth wooden slides to move quickly from one level to another. the speed of the miner at the bottom of the slide is much less than the calculated maximum possible speed. Explain why.

Answer 1

Step-by-step explanation:

The discrepancy between the calculated maximum possible speed of the miner and the actual speed observed at the bottom of the slide can be explained by several factors:

• Friction: Even though the wooden slides are smooth, there's still some degree of friction between the miner's clothing or equipment and the slide. This friction acts against the motion, slowing the miner down as they descend.
• Air Resistance: As the miner moves down the slide, they'll encounter air resistance, which increases with their speed. This resistance can play a significant role in reducing their final velocity

When analysing the motion of the miner on the slide, the primary principle at play is the conservation of energy. In an ideal situation without any external forces, the potential energy the miner has at the top of the slide is converted entirely into kinetic energy at the bottom:

`\[ PE_{\text{top}} = KE_{\text{bottom}} \]`

For potential energy
`\( PE \)` due to height
`\( h \)` and mass
`\( m \)`:

`\[ PE = mgh \]`

Where:

-
`\( g \)` is the acceleration due to gravity (approx.
`\( 9.81 \, \text{m/s}^2 \)`).

For kinetic energy
`\( KE \)` for a velocity
`\( v \)`:

`\[ KE = (1)/(2)mv^2 \]`

Equating the two energies:

`\[ mgh = (1)/(2)mv^2 \]`

From this, we can derive the theoretical maximum speed
`\( v \)` at the bottom:

`\[ v = √(2gh) \]`

However, the observed speed is less than this value, which brings us to the reason: friction.

The work done against friction is given by:

`\[ W = f * d \]`

Where
`\( f \)` is the force of friction and
`\( d \)` is the distance (length of the slide). The force of friction (assuming a constant coefficient of friction
`\( \mu \)`) is:

`\[ f = \mu * N \]`

Where
`\( N \)` is the normal (or perpendicular) force. For an inclined plane,
`\( N = m * g * \cos(\theta) \)`, where
`\( \theta \)` is the angle of the slide with respect to the horizontal.

Considering the work-energy theorem, the work done against friction reduces the kinetic energy of the miner:

`\[ (1)/(2)mv^2 = mgh - \mu * m * g * \cos(\theta) * d \]`

From this, you can derive the actual speed at the bottom considering friction. This speed will be less than the
`\( √(2gh) \)` derived earlier because of the energy lost to overcome the friction.

Air resistance (or drag) for objects moving through the air is given by:

`\[ F_{\text{drag}} = (1)/(2) * \rho * v^2 * A * C_d \]`

Where:

-
`\( \rho \)` is the air density (about
`\( 1.225 \, \text{kg/m}^3 \)` at sea level).

-
`\( v \)` is the velocity of the miner.

-
`\( A \)` is the cross-sectional area of the miner facing the direction of motion.

-
`\( C_d \)` is the drag coefficient, which depends on the shape of the object (for a human, this could be around 1.0-1.3 when facing the air, but this varies greatly).

The kinetic energy at the bottom, in the presence of both friction and air resistance, will be:

`\[ (1)/(2)mv^2 = mgh - \mu * m * g * \cos(\theta) * d - \int F_{\text{drag}} \, ds \]`

Where
`\( \int F_{\text{drag}} \, ds \)` represents the work done against air resistance over the distance of the slide.