Problem 21 - University Physics with Modern Physics

Question

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathrm{k}}.\) Use the work- energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{\mathrm{k}},\) and \(\alpha\).

Step-by-Step Solution

Answer

\( v = \sqrt{\frac{2gh}{\tan(\alpha) - \mu_k}} \)

Step 1: Analyze the forces acting on the box

Identify the forces acting on the box as it moves up the incline. The forces include gravity, normal force, and friction. The gravitational force can be split into components parallel and perpendicular to the incline. The frictional force acts in the opposite direction of motion and is calculated using the coefficient of friction and the normal force.

Step 2: Write the expression for energy conservation

According to the work-energy theorem, the work done by all forces equals the change in kinetic energy. The initial kinetic energy at the bottom of the incline is \( \frac{1}{2} mv^2 \). The final kinetic energy at the top is zero, as it just reaches the skier and stops. The gravitational potential energy gained is \( mgh \). The work done against friction \( W_f \) is \( f_d \cdot d \), where \( f_d = \mu_k mg \cos(\alpha) \) and \( d \) is the distance along the incline.

Step 3: Relate distance with height

Express the distance \( d \) along the incline in terms of the height \( h \) and the angle \( \alpha \). From trigonometry, \( \sin(\alpha) = \frac{h}{d} \). Thus, \( d = \frac{h}{\sin(\alpha)} \).

Step 4: Set up the equation using the work-energy principle

The initial kinetic energy plus the work done against friction must equal the gravitational potential energy at the top:\[\frac{1}{2} mv^2 + \mu_k mg \cos(\alpha) \cdot \frac{h}{\sin(\alpha)} = mgh\]

Step 5: Solve for initial speed \( v \)

Simplify the equation to solve for \( v \). Cancel \( m \) from both sides and rearrange terms:\[\frac{1}{2} v^2 = gh - \mu_k g \frac{h \cos(\alpha)}{\sin(\alpha)}\]Multiply by 2 and take the square root:\[v = \sqrt{2gh \left(1 - \frac{\mu_k \cos(\alpha)}{\sin(\alpha)}\right)}\]

Step 6: Express the final formula

The final formula for the minimum speed \( v \) you must give the box at the bottom of the incline is:\[ v = \sqrt{\frac{2gh}{\tan(\alpha) - \mu_k}} \]

Key Concepts

Kinetic Friction

Kinetic friction plays an essential role when any object moves over a surface. It's the force that opposes the motion, trying to slow the object down.
It depends on two key factors:

• The coefficient of kinetic friction \(\mu_k\), which is a measure of how "slippery" two surfaces are against each other.
• The normal force \(N\), which is the perpendicular force the surface exerts on the object. On an incline, this force is less than the object's weight since the incline reduces the direct contact.

The formula to calculate kinetic friction \(f_k\) is:
\[ f_k = \mu_k \cdot N \]
On an inclined plane, the normal force is given by \(N = mg \cos(\alpha)\), where \(\alpha\) is the angle of the incline. Therefore, kinetic friction can be summarized as:
\[ f_k = \mu_k mg \cos(\alpha) \]
Understanding this concept allows us to calculate the work done against friction as the box moves up the incline.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, different from a horizontal surface, typically used to move objects up or down with reduced force.
This angle, \(\alpha\), affects the way forces act on an object. Here's how it works:

• The gravitational force on the object can be divided into two components: parallel \(mg \sin(\alpha)\) and perpendicular \(mg \cos(\alpha)\) to the incline.
• The parallel component attempts to pull the object down the plane, while the frictional force opposes this motion.
• The perpendicular component impacts the normal force, which influences the friction.

On an inclined plane, the vertical distance \(h\) the box needs to overcome is related to the hypotenuse \(d\) of the incline by \(\sin(\alpha) = \frac{h}{d}\).
This formula is crucial for finding how far the box travels along the inclined plane, which impacts calculations of work and energy.

Potential Energy

Potential energy is the stored energy of an object due to its position relative to a gravitational field.
As you project the box up the incline, it gains gravitational potential energy, denoted as \(PE\).
The formula for gravitational potential energy is:

• \(PE = mgh\)

Where:

• \(m\) is the mass of the box.
• \(g\) is the acceleration due to gravity.
• \(h\) is the vertical height above the starting point.

The potential energy at the top of the incline balances the kinetic energy at the bottom minus the work done against friction while the box ascends. This energy transition is key to calculating the initial speed needed for the box to reach the skier at the top.

Initial Speed Calculation

Determining the initial speed \(v\) needed to project the box up the inclined plane involves the work-energy theorem.
This theorem states that the work done by all forces is equal to the change in kinetic energy of the object.
At the start:

• The box has initial kinetic energy \(\frac{1}{2}mv^2\).
• As it goes up, it loses speed until it reaches zero at the top, where all kinetic energy has turned into potential energy and work against friction.

Using the expression:
\[ \frac{1}{2} mv^2 + \mu_k mg \cos(\alpha) \cdot \frac{h}{\sin(\alpha)} = mgh \]
You solve for \(v\):
Cancel the mass \(m\) and rearrange the terms:
\( \frac{1}{2}v^2 = gh - \mu_k g \frac{h \cos(\alpha)}{\sin(\alpha)}\).
Multiply both sides by 2, and take the square root:
\[ v = \sqrt{rac{2gh}{\tan(\alpha) - \mu_k}} \]
With this formula, you can calculate the minimum speed needed for the box to overcome friction and gain enough potential energy to reach the skier.