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Chapter 6: Problem 12

A boxed \(10.0 \mathrm{~kg}\) computer monitor is dragged by friction \(5.50 \mathrm{~m}\) upward along a conveyor belt inclined at an angle of \(36.9^{\circ}\) above the horizontal. If the monitor's speed is a constant \(2.10 \mathrm{~cm} / \mathrm{s}\), how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Short Answer

Expert verified
The work done on the monitor by: (a) friction is 0 J, (b) gravity is -336.8 J, and (c) the normal force is 0 J.

Step by step solution

01

Identify Data

First, identify the data provided: the weight of the monitor (10.0 kg), the distance it's moved (5.50 m), the angle of the conveyor belt (36.9 degrees), and the constant speed it moves at (2.10 cm/s). Note that the speed must be converted to m/s for consistency in units. After conversion, we get a speed of 0.021 m/s.
02

Calculate Work Done by Friction

The work done by friction is zero because it is said the monitor's speed is a constant. That means there's no acceleration or deceleration, so there's no net external force doing work.
03

Calculate Work Done by Gravity

The work done by gravity is given by the equation \(W = F \times d \times cos(\Theta)\). The force \(F\) due to gravity here would be the weight of the monitor, which is mass \(\times\) acceleration due to gravity i.e, \(F = 10.0 kg \times 9.8 m/s^2 = 98 N\). The direction of this force is downwards, contrary to the direction of displacement, so the angle \(\Theta\) between the force and displacement is 180-36.9 = 143.1 degrees. After substituting all values, the work done by gravity comes out to be -336.8 J.
04

Calculate Work Done by Normal Force

The normal force is always perpendicular to the displacement of the monitor, thus it doesn't do any work on the monitor. The work done by the normal force is zero.

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Key Concepts

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

Work Done by Friction
When we talk about work done by friction in physics, we are referring to the force that opposes motion between two surfaces in contact. For an object moving at a constant speed, like the computer monitor in our exercise, frictional force must be present to balance out other forces and maintain the monitor's non-accelerating state. However, because the monitor maintains a constant velocity, the net work done by friction is zero. This might appear counterintuitive at first, but remember that work is defined as a force causing displacement. Since the net force (including friction) is not accelerating the monitor, there's no net work being done over the distance it travels.

Work Done by Gravity
Gravity is always at play, pulling objects towards the Earth's center with a force proportional to their mass. When calculating the work done by gravity, it's vital to note that this force acts downwards, which can be either aligned or against the direction of the object's displacement. In our exercise, since gravity acts downwards while the monitor is moved upward, the work done by gravity will be negative, indicating energy transfer in opposite direction to the displacement. Using the formula for work (\(W = F \times d \times \text{cos}(\theta)\)), we deduce that gravity does negative work on the monitor, encouraging a better understanding of energy conservation.

Normal Force Work
Delving into the concept of normal force work requires understanding the role of normal force in physics. Essentially, the normal force is the support force exerted by a surface perpendicularly to an object resting on it. It's a reactionary force preventing the object from moving into the surface. When an object moves along a surface without penetrating or lifting off the surface, as is the case with the monitor on the conveyor, the displacement is perpendicular to the normal force. Therefore, since work is only done when a force has a component along the direction of displacement, the normal force in our scenario does zero work on the monitor.

Kinetic Friction
Kinetic friction comes into play when objects are in motion. It acts to resist the relative motion of the objects in contact. Interestingly, kinetic friction depends on the normal force and the coefficient of kinetic friction, which is unique to the contacting materials. In our exercise, kinetic friction is what maintains the monitor's constant speed up the inclined plane, counteracting any tendency for acceleration due to other forces like gravity. Exploring this concept aids students in appreciating the intricate balance forces must strike to produce steady movement.

Inclined Plane Physics
Inclined plane physics involves understanding the forces and motion on a surface tilted at an angle to the horizontal. It's a fascinating concept that simplifies the study of forces in a two-dimensional plane. By analyzing forces in parallel and perpendicular components to the inclined plane, students can more readily solve a variety of problems. In our question, the monitor's movement up an inclined plane provides a practical example illustrating how forces like gravity and friction are altered when acting on an incline, as opposed to horizontal movement.

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Most popular questions from this chapter

A soccer ball with mass \(0.420 \mathrm{~kg}\) is initially moving with speed \(2.00 \mathrm{~m} / \mathrm{s}\). A soccer player kicks the ball, exerting a constant force of magnitude \(40.0 \mathrm{~N}\) in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to \(6.00 \mathrm{~m} / \mathrm{s} ?\)

A net horizontal force \(F\) is applied to a box with mass \(M\) that is on a horizontal, frictionless surface. The box is initially at rest and then moves in the direction of the force. After the box has moved a dis- tance \(D,\) the work that the constant force has done on it is \(W_{D}\) and the speed of the box is \(V\). The equation \(P=F v\) tells us that the instanta neous rate at which \(F\) is doing work on the box depends on the speed of the box. (a) At the point in the motion of the box where the force has done half the total work, and so has done work \(W_{D} / 2\) on the box that started from rest, in terms of \(V\) what is the speed of the box? Is the speed at this point less than, equal to, or greater than half the final speed? (b) When the box has reached half its final speed, so its speed is \(V / 2,\) how much work has been done on the box? Express your answer in terms of \(W_{D}\). Is the amount of work done to produce this speed less than, equal to, or greater than half the work \(W_{D}\) done for the full displacement \(D ?\)

A \(30.0 \mathrm{~kg}\) crate is initially moving with a velocity that has magnitude \(3.90 \mathrm{~m} / \mathrm{s}\) in a direction \(37.0^{\circ}\) west of north. How much work must be done on the crate to change its velocity to \(5.62 \mathrm{~m} / \mathrm{s}\) in a direction \(63.0^{\circ}\) south of east?

\(\mathrm{BIO}\) All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of object mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass \(70 \mathrm{~g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A \(70 \mathrm{~kg}\) athlete can maintain a power output of \(1.4 \mathrm{~kW}\) for no more than a few seconds; the steady power output of a typical athlete is only \(500 \mathrm{~W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

Varying Coefficient of Friction. A box is sliding with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\) on a horizontal surface when, at point \(P,\) it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at \(12.5 \mathrm{~m}\) past point \(P .\) (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of \(0.100 ?\)
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