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 \(\mu_k\). 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 \(g\), \(h\), \(\mu_k\), and \(\alpha\).

Short answer

The minimum speed needed is \( v_i = \sqrt{2gh (1 + \mu_k \cot(\alpha))} \).

Step-by-step solution

Apply Work-Energy Theorem

According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy. For the box, the change in kinetic energy (\( \Delta KE\) ) is given by \( \frac{1}{2} mv^2_f - \frac{1}{2} mv^2_i \). Since the box comes to rest at the top, \(v_f = 0\). Hence, \(\Delta KE = - \frac{1}{2} mv^2_i \).

Calculate Work Done Against Gravity

As the box moves up the incline, it gains gravitational potential energy equivalent to \( mgh \). This energy must be provided by the initial kinetic energy.

Calculate Work Done Against Friction

Friction does negative work on the box as it moves up the incline. The work done by friction \(W_f\) is given by \( \mu_k mg \cos(\alpha) d \), where \( d \) is the distance traveled along the incline. Use geometry to express \( d = \frac{h}{\sin(\alpha)} \), thus \(W_f = \mu_k mg \frac{h \cos(\alpha)}{\sin(\alpha)}\).

Equate Work-Energy Balance

Set up the equation for the work-energy balance: \[0 = - \frac{1}{2} mv_i^2 + mgh + \mu_k mg \frac{h \cos(\alpha)}{\sin(\alpha)}\].

Solve for Initial Velocity

Rearrange the equation to solve for the initial velocity \( v_i \): \[ \frac{1}{2} mv_i^2 = mgh (1 + \mu_k \cot(\alpha)) \].Solve for \( v_i \): \[ v_i = \sqrt{2gh (1 + \mu_k \cot(\alpha))} \].

Key concepts

Gravitational Potential Energy

Gravitational potential energy is a form of energy associated with the height of an object. When you lift an object against the force of gravity, you're doing work on it, increasing its potential to do work when it falls back down. For our exercise, as the box of supplies is projected up an incline, it moves to a greater height, thereby increasing its gravitational potential energy.
This energy is calculated using the formula:

• \( mgh \)
• where \( m \) is the mass of the box, \( g \) is the acceleration due to gravity, and \( h \) is the vertical distance covered.

In this scenario, the initial kinetic energy imparted must be enough to allow the box to reach the desired height by overcoming both gravity and friction.

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. Unlike static friction, which prevents an object from moving, kinetic friction acts on objects that are already in motion. The force of kinetic friction can be calculated using the formula:

• \( f_k = \mu_k N \)
• where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.

In our inclined plane problem, kinetic friction works against the box as it slides up the slope, doing negative work by taking away some of the box's energy. This force depends on the box's weight and the angle of the incline, which affects the normal force experienced by the box.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, and it's a classic physics scenario used to analyze forces and energy. In this exercise, the incline not only affects how gravity pulls on the box but also the normal force impacting kinetic friction. The angle \( \alpha \) affects how much of the gravitational force acts parallel to the plane.
The characteristics of the inclined plane are crucial as they determine how much work is done against gravity and friction. The geometry of the plane helps in:

• Computing the distance \( d \) using \( d = \frac{h}{\sin(\alpha)} \), which is key to calculating the work done.
• Defining the components of forces acting on the box.

Initial Velocity

Initial velocity is the speed given to an object to start its motion. For this problem, determining the required initial velocity is essential to ensure the box reaches the skier. The initial speed must overcome the energy losses due to gravitational potential energy gain and kinetic friction.
Using the work-energy theorem, we find that the minimum initial velocity \( v_i \) needed is: \[ v_i = \sqrt{2gh (1 + \mu_k \cot(\alpha))} \]This equation results from balancing the energies involved:

• The loss in kinetic energy as the box climbs.
• The increase in potential energy.
• The work done against friction.

The initial velocity ensures that the starting kinetic energy is sufficient for the box to overcome all these energy demands and safely reach its target.