/*! 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 block of mass \(m\) is release... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 62

A block of mass \(m\) is released from rest at the top of an incline that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction between the block and incline is \(\mu_{k}\). The top of the incline is a vertical distance \(h\) above the bottom of the incline. Derive an expression for the work \(W_{f}\) done on the block by friction as it travels from the top of the incline to the bottom. When \(\alpha\) is decreased, does the magnitude of \(W_{f}\) increase or decrease?

Short Answer

Expert verified
The work done by the friction on the block as it slides down the incline can be given by the expression \(W_{f} = - \mu_{k} mg\cos(\alpha) * \frac{h}{\sin(\alpha)}\). If the incline angle, \(\alpha\), is decreased, the magnitude of the work done by friction, \(W_{f}\), would increase.

Step by step solution

01

Understanding the Forces

Identify the forces acting on the block. The forces acting on the block are gravity, normal force, and friction. Gravity's force on the block can be written as \(mg\) (mass times the acceleration due to gravity) and directed downwards, the normal force perpendicular to the surface of the incline and the friction force opposite to the direction of motion. The frictional force is given by the equation \(F_{f}= \mu_{k} F_{n}\) where \(F_{f}\) is the frictional force, \(\mu_{k}\) is the kinetic coefficient of friction and \(F_{n}\) is the normal force.
02

Calculate the Normal Force

The normal force on an incline is given by \(F_{n} = mg\cos(\alpha)\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(\alpha\) is the angle of the incline. Here the normal force is not equal to weight since it is on an inclined plane.
03

Calculate the Frictional Force

Substitute the normal force from step 2 into the frictional force formula from step 1. This gives \(F_{f} = \mu_{k} mg\cos(\alpha)\). This is the frictional force experienced by the block as it slides down the incline.
04

Calculate the work done by Friction

The work done by a constant force is given by \(W = Fd\cos(\theta)\), where \(F\) is the force, \(d\) is the displacement and \(\theta\) is the angle between the force and the displacement. In this case, friction is opposite to the direction of displacement, making \(\theta = 180^{\circ}\) or \(\pi\) rad and cosine of 180 is -1. The displacement \(d\) is the length of the incline which can be found by \(d = \frac{h}{\sin(\alpha)}\). This gives the work done by friction \(W_{f} = - \mu_{k} mg\cos(\alpha) * \frac{h}{\sin(\alpha)}\)
05

Interpretation of the Work

If we decrease the incline angle, \(\alpha\), we can see from the derived expression in step 4 that the magnitude of the work done by friction, \(W_{f}\), would increase. This occurs because decreasing the angle increases the distance the block has to travel, thereby increasing the work done by friction.

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Key Concepts

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

Work Done by Friction
Friction is a force that always acts to oppose motion, and as such, it produces work that is done against the motion of an object. In the scenario of a block sliding down an inclined plane, the frictional force is in the opposite direction of the block's movement.
When the block slides down, the work done by friction can be calculated using the formula:
  • Work ( \( W \) ) = Force ( \( F \) ) x Displacement ( \( d \) ) x Cosine of the angle ( \( \theta \) ).
In this case, the angle \( \theta \) is 180掳 because the frictional force direction is opposite to the displacement direction.
Consequently, the cosine of 180掳 is -1, which indicates the work done by friction acts to remove energy from the system, effectively slowing the block down. The displacement (\( d \)) is calculated along the plane, which is crucial for understanding how friction affects the block's motion over a distance.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It's used to analyze how forces act on objects placed on it. When examining the motion of a block on an incline, understanding how the components of gravitational force interact with the surface is essential.
The gravitational force \( mg \) acts downward, but only part of it affects the block's motion along the plane.
  • The component parallel to the plane is responsible for the sliding, calculated as \( mg\sin(\alpha) \) .
  • The component perpendicular to the plane, which is \( mg\cos(\alpha) \) , determines the normal force.
The angle \( \alpha \) not only affects these force components but also the distance the block travels. A smaller angle elongates this distance, changing both the speed and work done by friction dramatically.
Kinetic Coefficient of Friction
The kinetic coefficient of friction (\( \mu_{k} \)) is a dimensionless number that quantifies the frictional force between moving surfaces. Its value depends on the materials in contact and is crucial in determining the frictional force on an incline.
This coefficient allows us to compute the frictional force (\( F_{f} \)) using:
  • \( F_{f} = \mu_{k} F_{n} \) , where \( F_{n} \) is the normal force.
On an inclined plane, \( F_{n} \) is calculated as \( mg\cos(\alpha) \), which adjusts the frictional force experienced by the object.
A higher \( \mu_{k} \) means greater resistance to movement, leading to more energy lost as the block slides down the incline. Understanding this helps to predict how variations in the surface and angle of the incline affect the motion and work done by friction as the block descends.

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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 \(0.200 \mathrm{~m} ?\)

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 \(g, h, \mu_{k},\) and \(\alpha\)

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

You and your bicycle have combined mass \(80.0 \mathrm{~kg}\). When you reach the base of a bridge, you are traveling along the road at \(5.00 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.74\) ). At the top of the bridge, you have climbed a vertical distance of \(5.20 \mathrm{~m}\) and slowed to \(1.50 \mathrm{~m} / \mathrm{s}\). Ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

One end of a horizontal spring with force constant \(130.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical wall. A \(4.00 \mathrm{~kg}\) block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is \(\mu_{\mathrm{k}}=0.400 .\) You apply a constant force \(\vec{F}\) to the block. \(\vec{F}\) has magnitude \(F=82.0 \mathrm{~N}\) and is directed toward the wall. At the instant that the spring is compressed \(80.0 \mathrm{~cm},\) what are (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?
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