/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 You are a member of an Alpine Re... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mu_{\mathrm{k}}.\) 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 \(\alpha\).

Short Answer

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

Step by step solution

01

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.
02

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.
03

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, \( d = \frac{h}{\sin(\alpha)} \).
04

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:\[\frac{1}{2} mv^2 + \mu_k mg \cos(\alpha) \cdot \frac{h}{\sin(\alpha)} = mgh\]
05

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:\[v = \sqrt{2gh \left(1 - \frac{\mu_k \cos(\alpha)}{\sin(\alpha)}\right)}\]
06

Express the final formula

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

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Key Concepts

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

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.

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Most popular questions from this chapter

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

Animal Energy. BIO Adult cheetahs, the fastest of the great cats, have a mass of about 70 \(\mathrm{kg}\) and have been clocked running at up to 72 \(\mathrm{mph}(32 \mathrm{m} / \mathrm{s})\) . (a) How many joules of kinetic energy does such a swift cheetah have? (b) By what factor would its kinetic energy change if its speed were doubled?

CALC An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2}\) . (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1},\) is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2} .\) How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

CPA small block with a mass of 0.0900 \(\mathrm{kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.75). The block is originally revolving at a distance of 0.40 \(\mathrm{m}\) from the hole with a speed of 0.70 \(\mathrm{m} / \mathrm{s} .\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 \(\mathrm{m} .\) At this new distance, the speed of the block is observed to be 2.80 \(\mathrm{m} / \mathrm{s}\) . (a) What is the tension in the cord in the original situation when the block has speed \(v=0.70 \mathrm{m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation when the block has speed \(v=2.80 \mathrm{m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h,\) making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\) . (b) Use the work-energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if it had been dropped from height \(h,\) independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy friction- less hill, starting from rest 15.0 m above the bottom.

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