/*! 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 62 A \(20.0 \mathrm{~kg}\) box is p... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(20.0 \mathrm{~kg}\) box is pulled along a rough horizontal surface with a rope at a constant speed of \(8 \mathrm{~m} / \mathrm{s}\). The rope is parallel to the floor, and the coefficient of kinetic friction between the box and the floor is 0.200 . What power must the rope supply to the box? Where does this energy go?

Short Answer

Expert verified
The rope supplies 313.6 W to the box, dissipated as thermal energy.

Step by step solution

01

Understand the Forces Involved

Since the box is moving at a constant speed, the net force on it is zero. This means that the tension in the rope providing the pulling force is equal to the frictional force opposing the motion.
02

Calculate the Frictional Force

The frictional force \( F_{friction} \) can be calculated using the formula \( F_{friction} = \mu_k \times F_{normal} \), where \( \mu_k \) is the coefficient of kinetic friction and \( F_{normal} = m \times g \) is the normal force. With \( \mu_k = 0.200 \), mass \( m = 20.0 \mathrm{~kg} \), and \( g = 9.8 \mathrm{~m/s^2} \), we have: \[ F_{friction} = 0.200 \times 20.0 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} = 39.2 \mathrm{~N}. \]
03

Determine the Power Required

Power is defined as the work done over time, or alternatively, the force multiplied by the velocity. Since the friction force is opposing the motion, the same amount of force must be supplied by the rope to maintain constant speed. Therefore, the power \( P \) required is given by \[ P = F_{friction} \times v = 39.2 \mathrm{~N} \times 8 \mathrm{~m/s} = 313.6 \mathrm{~W}. \]
04

Explain the Energy Transfer

The energy supplied by the rope is used to overcome the friction between the box and the floor. This energy is dissipated as thermal energy due to friction, warming up both the box and the floor slightly.

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Key Concepts

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

Kinetic Friction in Physics
Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is a result of the interactions between the microscopic bumps and grooves on the surfaces in contact. To calculate kinetic friction, we use the formula:
  • \( F_{friction} = \mu_k \times F_{normal} \)
where \( \mu_k \) is the coefficient of kinetic friction and \( F_{normal} \) is the normal force, which is the perpendicular force that the surface exerts on the object.
The normal force is usually equal to the force of gravity on the object (\( m \times g \), where \( m \) is mass and \( g \) is acceleration due to gravity). This means that heavier objects or those with a higher coefficient of kinetic friction experience more frictional force.
In our case, the rope must supply a force equal to this frictional force to keep the box moving at a constant speed. This means that all the energy used to pull the box is transferred to overcoming kinetic friction.
Understanding Force and Motion
In physics, force and motion are fundamental concepts. Force is any interaction that, when unopposed, changes the motion of an object. Motion is the change in position of an object over time. When a box is pulled on a surface, a balance of forces occurs.
Here, we are dealing with constant speed, indicating that the net force acting on the box is zero. This means that the pulling force exerted by the rope and the kinetic frictional force are equal and opposite.
Mathematically, we express this as:
  • \( F_{pull} = F_{friction} \)
This equilibrium is crucial for maintaining the motion at a constant speed, demonstrating an essential principle in ... Newton's first law of motion: an object will remain at rest or in uniform motion unless acted upon by a net external force.
Energy Dissipation
Energy dissipation refers to the process of energy being transformed from one form to another, typically resulting in energy loss in the system, often as heat. When the box is dragged over a rough surface, energy is continually expended to overcome the frictional forces.
The power provided by the rope translates into mechanical work, which is then converted primarily into thermal energy due to friction.
This conversion is represented in the equation:
  • \( P = F_{friction} \times v \)
where \( P \) is the power needed to sustain the box's motion.
Heat production due to friction not only warms the contacting surfaces but also effectively dissipates the energy, highlighting the principle of energy conservation where energy is "lost" from the system as unusable heat. This is a natural consequence of the second law of thermodynamics, where energy tends to disperse from high concentration regions to the surroundings.

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Most popular questions from this chapter

While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0 \mathrm{~N}\) toolbox, causing it to start sliding downward, starting from rest. If it starts \(4.25 \mathrm{~m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is \(22.0 \mathrm{~N} ?\)

A \(62.0 \mathrm{~kg}\) skier is moving at \(6.50 \mathrm{~m} / \mathrm{s}\) on a frictionless, horizontal snow-covered plateau when she encounters a rough patch \(3.50 \mathrm{~m}\) long. The coefficient of kinetic friction between this patch and her skis is 0.300 . After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill \(2.50 \mathrm{~m}\) high. (a) How fast is the skier moving when she gets to the bottom of the hill? (b) How much internal energy was generated in crossing the rough patch?

A tennis player hits a \(58.0 \mathrm{~g}\) tennis ball so that it goes straight up and reaches a maximum height of \(6.17 \mathrm{~m} .\) How much work does gravity do on the ball on the way up? On the way down?

A spring with spring constant \(100 \mathrm{~N} / \mathrm{m}\) and unstretched length \(0.4 \mathrm{~m}\) has one end anchored to a wall and a force \(F\) is applied to the other end. If the force \(F\) does \(200 \mathrm{~J}\) of work in stretching out the spring, (a) what is its final length, (b) what is the magnitude of \(F\) at maximum elongation?

Stopping distance of a car. The driver of an \(1800 \mathrm{~kg}\) car (including passengers) traveling at \(23.0 \mathrm{~m} / \mathrm{s}\) slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically \(0.700 .\) (a) Use the work-energy theorem to calculate how far the car will travel before stopping. (b) How far would the car travel if it were going twice as fast? (c) What happened to the car's original kinetic energy?

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