/*! 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 A cord is used to vertically low... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 21

A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

Short Answer

Expert verified
(a) - \( M\times\frac{3g}{4} \times d \), (b) \( Mg \times d \), (c) \( \frac{Mg}{4} \times d \), (d) \( \sqrt{\frac{gd}{2}} \).

Step by step solution

01

Determine the Acceleration of the Block

Given that the block experiences a constant downward acceleration of \( \frac{g}{4} \), where \( g \) is the acceleration due to gravity, the net acceleration \( a \) is \( a = \frac{g}{4} \). This means the total force acting downward on the block is \( M \times \frac{g}{4} \).
02

Identify the Forces Acting on the Block

There are two forces acting on the block: gravitational force \( F_{gravity} = Mg \) acting downward and tension in the cord \( F_{cord} \). These forces result in the block鈥檚 net downward acceleration of \( \frac{g}{4} \). Thus, we have: \[Mg - F_{cord} = M\times\frac{g}{4}\]
03

Solving for Cord's Force

Rearrange the equation fro Step 2:\[F_{cord} = Mg - M\times\frac{g}{4} = M\left(g - \frac{g}{4}\right) = M\times\frac{3g}{4}.\]Thus, the force exerted by the cord is \( M\times\frac{3g}{4} \).
04

Work Done by the Cord's Force

The work done by the cord's force can be calculated using the formula: \\[\text{Work}_{cord} = F_{cord} \times d \times \cos(180^\circ) = - M\times\frac{3g}{4} \times d.\] \The work is negative since the force and displacement are in opposite directions.
05

Work Done by Gravitational Force

The work done by the gravitational force is given by the product of the force and the distance moved in the direction of the force: \[\text{Work}_{gravity} = Mg \times d.\]
06

Calculate the Kinetic Energy of the Block

By the work-energy principle, the total work done on the block is equal to its change in kinetic energy:\[\Delta KE = \text{Work}_{cord} + \text{Work}_{gravity} = - M\times\frac{3g}{4} \times d + Mg \times d.\]Simplifying, we find,\[\Delta KE = M\times\frac{g}{4} \times d.\]So, the kinetic energy \( KE = \frac{Mg}{4} \times d \).
07

Determine the Speed of the Block

The kinetic energy \( KE \) is also given by \( \frac{1}{2} Mv^2 \). We can set this equal to the equation found in Step 6:\[\frac{1}{2} M v^2 = \frac{Mg}{4} \times d.\]Solve for \( v \):\[v^2 = \frac{2g}{4} \times d \] \[v = \sqrt{\frac{gd}{2}}.\]

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in mechanics that connects the ideas of work and energy. It states that the total work done on an object is equivalent to the change in its kinetic energy. This idea helps us understand how forces acting on an object influence its motion.
When we say 'work is done,' it means a force causes the object to move. As an object moves, it gains or loses energy depending on the direction of the force. In simple terms, if the total work done is positive, the object speeds up, gaining kinetic energy. Conversely, if the work done is negative, the object slows down, losing kinetic energy.
In our exercise, the work done by the cord and the gravitational force together determined the total change in kinetic energy. Even though the cord鈥檚 force does negative work (acting against the motion), the gravitational force does positive work (aiding in the motion), which ultimately leads to the object gaining kinetic energy.
Gravitational Force
Gravitational force is a universal attraction force that acts between masses. On Earth, this force provides a constant acceleration, typically denoted by the symbol \( g \), approximately equal to \( 9.8 \text{ m/s}^2 \). Any object that is free to move in response to the gravitational force will accelerate downwards at this rate.
In the original exercise, the gravitational force acts on a block with a magnitude of \( Mg \), pointing downward. This force contributes positively to the block's motion, helping it move downwards.
  • Gravitational Force Formula: \( F_{gravity} = Mg \). This is the product of the object's mass \( M \) and the gravity \( g \).
  • Work Done by Gravity: The work done by this force over a distance \( d \) is calculated by \( ext{Work}_{gravity} = Mg imes d \), since it acts in the direction of motion.
This concept not only impacts the motion of the block but also plays a role in the calculation of the work-energy principle by contributing to the total work done on the object.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving possesses kinetic energy, and this energy increases with the speed of the object. The amount of kinetic energy an object has depends on its mass and velocity.
This energy is calculated using the formula:
  • Kinetic Energy Formula: \( KE = \frac{1}{2} M v^2 \)
In the context of our exercise, after falling a distance \( d \), the block gains kinetic energy. Initially, since the block was stationary, its kinetic energy was zero. As work is done on the block, this energy increases.
The solution shows that the new kinetic energy, after accounting for all forces, is \( \frac{Mg}{4} imes d \). This value is obtained from the net work done on the block, highlighting how kinetic energy stems from forces acting during motion.

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To push a \(25.0 \mathrm{~kg}\) crate up a frictionless incline, angled at \(25.0^{\circ}\) to the horizontal, a worker exerts a force of \(209 \mathrm{~N}\) parallel to the incline. As the crate slides \(1.50 \mathrm{~m},\) how much work is done on the crate by (a) the worker's applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

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