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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

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

Step by step solution

01

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, \(\Delta KE = - \frac{1}{2} mv^2_i \).
02

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.
03

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 \(W_f = \mu_k mg \frac{h \cos(\alpha)}{\sin(\alpha)}\).
04

Equate Work-Energy Balance

Set up the equation for the work-energy balance: \[0 = - \frac{1}{2} mv_i^2 + mgh + \mu_k mg \frac{h \cos(\alpha)}{\sin(\alpha)}\].
05

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 \( v_i \): \[ v_i = \sqrt{2gh (1 + \mu_k \cot(\alpha))} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work\(-\)energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k = 450\) N/m and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 kg and is to reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 \(\times\) 10\(^6\) N, one 14\(^\circ\) west of north and the other 14\(^\circ\) east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

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