/*! 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 63 A 0.150-kg block of ice is place... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.150-kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 m above the floor. The spring has force constant 1900 N/m and is initially compressed 0.045 m. The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

Short Answer

Expert verified
The speed of the block when it reaches the floor is 7.01 m/s.

Step by step solution

01

Calculate Potential Energy in Compressed Spring

The initial potential energy in the spring when compressed can be calculated using the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2} k x^2 \] where \( k = 1900 \, \text{N/m} \) is the spring constant and \( x = 0.045 \, \text{m} \) is the compression distance. Plug in the values: \[ PE_{spring} = \frac{1}{2} \times 1900 \times (0.045)^2 \] \[ PE_{spring} = 1.92 \, \text{J} \].
02

Calculate Kinetic Energy as Block Leaves the Table

Assuming negligible friction, the potential energy stored in the spring is converted entirely into the kinetic energy of the block as it leaves the table. Thus, we can set \( KE = PE_{spring} \). Using the kinetic energy formula: \[ KE = \frac{1}{2} m v^2 \] where \( m = 0.150 \, \text{kg} \) is the mass of the ice block. Equate and solve for \( v \): \[ \frac{1}{2} \times 0.150 \times v^2 = 1.92 \] \[ v^2 = \frac{1.92}{0.075} \] \[ v^2 = 25.6 \] \[ v = 5.06 \, \text{m/s} \].
03

Calculate Velocity of the Block Upon Reaching the Floor

As the block falls from the table, its velocity increases due to gravitational acceleration. Calculate the additional speed gained using: \[ v_f^2 = v^2 + 2gh \] where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 1.20 \, \text{m} \) is the height of the table. Plug in the values: \[ v_f^2 = (5.06)^2 + 2 \times 9.81 \times 1.20 \] \[ v_f^2 = 25.6 + 23.544 \] \[ v_f^2 = 49.144 \] \[ v_f = 7.01 \, \text{m/s} \].

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Key Concepts

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

Potential Energy
Potential energy is the stored energy of an object due to its position or configuration. In the context of springs, we talk about elastic potential energy. Elastic potential energy is stored in a compressed or extended spring, ready to be converted into kinetic energy. For a compressed or stretched spring, the potential energy can be calculated using the formula:
\[PE_{spring} = \frac{1}{2} k x^2 \]where:
  • \( k \) is the spring constant, a measure of the stiffness of the spring.
  • \( x \) is the distance the spring is compressed or extended.
In our problem, the spring has a constant \( k = 1900 \, \text{N/m} \) and is compressed \( 0.045 \, \text{m} \). Doing the math gives us the potential energy stored, \( PE_{spring} = 1.92 \, \text{J} \). This energy is crucial as it later transforms into kinetic energy when released.
Kinetic Energy
Kinetic energy is the energy of motion. Once the potential energy in a spring is released, it is largely converted into the kinetic energy of the moving object. The formula for kinetic energy is:
\[KE = \frac{1}{2} m v^2\]where:
  • \( m \) is the mass of the object.
  • \( v \) is the velocity of the object.
In the scenario given, the potential energy of the spring is transformed into the kinetic energy of the ice block as it leaves the table. By assuming negligible friction, we make the calculation straightforward by setting \( KE = PE_{spring} \). This allows us to solve for velocity \( v \) of the block, which turns out to be \( 5.06 \, \text{m/s} \). This velocity is before it starts gaining additional speed due to gravity.
Gravitational Acceleration
Gravitational acceleration is the increase in speed of an object as it is pulled towards the Earth. When the ice block leaves the edge of the table, it is subject to the force of gravity, with an acceleration of \( g = 9.81 \, \text{m/s}^2 \). This force causes the block to gain velocity as it descends.The effect of gravity on falling objects can be mathematically expressed through the kinematic equation:
\[v_f^2 = v^2 + 2gh\]where:
  • \( v \) is the initial velocity when the block leaves the table.
  • \( g \) is the gravitational acceleration.
  • \( h \) is the height from which the object falls.
For the block of ice, it starts with an initial velocity \( v = 5.06 \, \text{m/s} \) from the spring release and gains additional speed as it falls from a height of \( 1.20 \, \text{m} \). Solving the equation reveals its impact speed as \( 7.01 \, \text{m/s} \). This showcases gravitational potential energy converting to kinetic energy as it falls.
Spring Mechanics
Spring mechanics involve the principles of force and energy as applied to springs. Springs store energy when compressed or stretched and release this energy as they return to their original form. The two main types of energy involved in spring mechanics are potential energy and kinetic energy.The spring constant \( k \) is fundamental in determining how much a spring will compress or extend under a given force. It is essentially a measure of stiffness:
  • A higher \( k \) means a stiffer spring.
  • A lower \( k \) means a more flexible spring.
The relation between force, spring constant, and displacement is expressed by Hooke's Law:
\[F = -kx\]where:
  • \( F \) is the force applied.
  • \( x \) is the displacement from equilibrium.
In our problem, once the spring is released, it transfers potential energy into kinetic energy, which propels the ice block. This conversion highlights the efficiency of springs in energy transformation. Understanding spring mechanics helps us appreciate how everyday items like mattresses and car suspensions work effectively.

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Most popular questions from this chapter

A 2.50-kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m. The spring has force constant 840 N/m. The coefficient of kinetic friction between the floor and the block is \(\mu_k =\) 0.40. The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)

The \(food \, calorie\), equal to 4186 J, is a measure of how much energy is released when the body metabolizes food. A certain fruitandcereal bar contains 140 food calories. (a) If a 65kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only 20% of the food calories go into mechanical energy, what would be the answer to part (a)? (\(Note\): In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.)

A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A small block with mass 0.0500 kg slides in a vertical circle of radius \(R =\) 0.800 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45\(^\circ\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of 30\(^\circ\) with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. Ignore air resistance and the mass of the vine.

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