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A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20 .\) What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

Short Answer

Expert verified
(a) The work done by the force is 1979.28 J. (b) The increase in thermal energy is 509.28 J.

Step by step solution

01

Understanding the Forces

To solve for the work done, we need to understand the forces acting on the trunk as it moves up the incline. The trunk is subject to gravity, normal force, kinetic friction, and the applied force. Since it's moving at constant speed, the net force along the incline is zero.
02

Calculate the Normal Force

The normal force (\(F_{N}\)) is the component of gravity perpendicular to the incline. It is calculated as:\[ F_{N} = mg \cos(\theta) \]where:- \(m = 50 \text{ kg}\)- \(g = 9.8 \text{ m/s}^2\)- \(\theta = 30^\circ\)Substitute these values:\[ F_{N} = 50 \times 9.8 \times \cos(30^\circ) = 50 \times 9.8 \times 0.866 = 424.38 \text{ N} \]
03

Calculate the Kinetic Friction Force

The kinetic friction force (\(F_{f}\)) is given by:\[ F_{f} = \mu F_{N} \]where \(\mu = 0.20\).Substitute \(F_{N} = 424.38 \; \mathrm{N}\):\[ F_{f} = 0.20 \times 424.38 = 84.88 \text{ N} \]
04

Calculate the Component of Gravitational Force Down the Incline

The component of the gravitational force down the incline (\(F_{g,\text{down}}\)) is:\[ F_{g,\text{down}} = mg \sin(\theta) \]Substitute the known values:\[ F_{g,\text{down}} = 50 \times 9.8 \times \sin(30^\circ) = 50 \times 9.8 \times 0.5 = 245 \text{ N} \]
05

Determine the Applied Force

Since the net force is zero for constant speed, the applied force (\(F_{a}\)) must balance both the kinetic friction and the gravitational component:\[ F_{a} = F_{f} + F_{g,\text{down}} \]\[ F_{a} = 84.88 + 245 = 329.88 \text{ N} \]
06

Calculate the Work Done by the Applied Force

The work done by the applied force (\(W_{a}\)) is the product of the applied force and the distance along the incline:\[ W_{a} = F_{a} \times d \]where \(d = 6.0 \text{ m}\).\[ W_{a} = 329.88 \times 6.0 = 1979.28 \text{ J} \]
07

Determine the Increase in Thermal Energy

The increase in thermal energy is equal to the work done against friction. Since the trunk moves at constant speed, all the work done by friction translates directly to thermal energy:\[ \Delta E_{\text{thermal}} = F_{f} \times d \]Substitute the values:\[ \Delta E_{\text{thermal}} = 84.88 \times 6.0 = 509.28 \text{ J} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that opposes the relative motion of surfaces in contact as one slides over the other. In the case of the trunk being pulled up the incline, kinetic friction acts as a resisting force against the movement. This type of friction is characterized by a coefficient, denoted as \( \mu \), which relates the friction force to the normal force. In our exercise, the coefficient of kinetic friction is given as \(0.20\).

The kinetic friction force \( (F_{f}) \) can be calculated using the formula:
  • \( F_{f} = \mu F_{N} \)
  • \( F_{N} \) is the normal force, which in this context is the component of the gravitational force perpendicular to the inclined surface.
The normal force here ensures that the trunk remains on the path of the incline, but it doesn't move upwards or downwards off the incline itself. Once you know the normal force, calculating the kinetic friction becomes straightforward, as seen in the step-by-step solution, where \( F_{f} \) was found to be \( 84.88 \text{ N} \). This force is crucial as it influences both the work done by the applied force and also the energy balance in this scenario.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, other than 90 degrees, to the horizontal. This simple machine helps in lifting or moving objects with less effort than needed on a perfectly vertical lift. Understanding the forces at play on an inclined plane is key to solving problems such as our trunk exercise.

In this scenario, the trunk is being pulled up a \(30^\circ\) incline. The forces acting on the trunk include gravitational force, normal force, applied force, and kinetic friction. These forces can be broken down into components that are parallel and perpendicular to the plane. Because the question specifies a constant speed, the net force on the trunk is zero. This means the applied force has to exactly counteract both the gravitational pull down the slope and the opposing kinetic friction force. In our solution, the applied force needed was calculated to be \( 329.88 \text{ N} \).

Inclined planes nicely demonstrate how physics and mathematics can reduce effort, allowing us to move items efficiently. It's important to decompose forces correctly to get accurate results and insights into the mechanics at play.
Thermal Energy
Thermal energy is the energy that arises from the motion of particles in a substance when two surfaces are in contact, such as in our trunk problem. As the trunk slides up the incline, kinetic friction does work against the motion, and this friction generates heat, which increases the system's thermal energy.

In the trunk problem, the work done to overcome friction results in an increase in thermal energy, as all of this work is converted into heat due to the resisting nature of friction. The amount of thermal energy created can be calculated using the work equation for friction:
  • \( \Delta E_{\text{thermal}} = F_{f} \times d \)
  • Here, \( F_{f} = 84.88 \text{ N} \) and \( d = 6.0 \text{ m} \).
This gives a total increase in thermal energy of \( 509.28 \text{ J} \).

Understanding these energetics is crucial in practical applications, such as engineering and safety mechanisms, where controlling heat generation and dissipation is essential for efficiency and function. This knowledge helps in designing systems that manage energy effectively, minimizing unwanted heat that can otherwise wear down systems and increase maintenance needs.

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Most popular questions from this chapter

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. \(8-37\) ) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

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A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

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