/*! 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 69 A \(0.150-\mathrm{kg}\) block of... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m} .\) 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 of ice when it reaches the floor is approximately 1.41 m/s.

Step by step solution

01

Identify Known Values and Variables

We are given the mass of the block \( m = 0.150\,\text{kg} \), the spring constant \( k = 1900\,\text{N/m} \), the initial compression of the spring \( x = 0.045\,\text{m} \), and the height \( h = 1.20\,\text{m} \). We want to find the speed of the block when it reaches the floor, denoted as \( v_f \).
02

Calculate Initial Elastic Potential Energy

The potential energy stored in the spring when compressed is given by the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2}kx^2 \]Substitute the given values: \[ PE_{spring} = \frac{1}{2} \times 1900 \times (0.045)^2 = 1.9125\,\text{J} \]
03

Calculate Gravitational Potential Energy Before Block Falls

The gravitational potential energy of the block when at the height \( h \) is calculated as: \[ PE_{gravity} = mgh \]Substitute the values: \[ PE_{gravity} = 0.150 \times 9.8 \times 1.20 = 1.764\,\text{J} \]
04

Conservation of Mechanical Energy

Since there is negligible friction, we can assume conservation of mechanical energy. The total mechanical energy at the start (elastic potential energy) equals the kinetic energy and gravitational potential energy when the block hits the floor: \[ PE_{spring} = KE + PE_{gravity} \]
05

Calculate Kinetic Energy When Block Reaches Floor

Rearrange the conservation equation to solve for the kinetic energy:\[ KE = PE_{spring} - PE_{gravity} = 1.9125 - 1.764 = 0.1485\,\text{J} \]
06

Solve for the Speed Using Kinetic Energy

The kinetic energy of the block when it hits the floor is given by \[ KE = \frac{1}{2}mv^2 \]Set this equal to the kinetic energy found in Step 5 and solve for \( v \):\[ 0.1485 = \frac{1}{2} \times 0.150 \times v^2 \]\[ v^2 = \frac{2 \times 0.1485}{0.150} \approx 1.98 \]\[ v \approx \sqrt{1.98} \approx 1.41\,\text{m/s} \]

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Key Concepts

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

Elastic Potential Energy
When you compress or stretch a spring, you store energy in the form of elastic potential energy. This energy can do work, like pushing or pulling an object, when the spring returns to its original shape. Elastic potential energy is calculated using the formula:
  • \( PE_{spring} = \frac{1}{2}kx^2 \)
Where \( k \) is the spring constant (a measure of the spring's stiffness) and \( x \) is the compression or extension distance from its natural length.
This equation tells us how much energy is "stored" in a spring by either compressing or stretching it. In the exercise, the spring was initially compressed 0.045 meters contributing energy to launch the block of ice across the table.
Kinetic Energy
Kinetic energy is the energy of motion. If an object is moving, it has kinetic energy. It is given by the equation:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( m \) is the mass of the object, and \( v \) is its velocity. This equation emphasizes how kinetic energy depends on both the mass of the object and its speed.
The faster something moves, the more kinetic energy it has, and the same goes if the object's mass increases. This concept was crucial in our exercise to determine how fast the ice block was moving once it reached the floor after falling off the table.
Gravitational Potential Energy
Gravitational potential energy is energy an object possesses because of its position relative to the Earth. The higher an object is, the more gravitational potential energy it has. It is calculated as:
  • \( PE_{gravity} = mgh \)
where \( m \) is the mass, \( g \) is the acceleration due to gravity \( (9.8 \ \text{m/s}^2) \), and \( h \) is the height above the ground.
In our problem, the block of ice initially starts with gravitational potential energy due to its height on the tabletop. This energy shifts into kinetic energy as it falls, with the energy transformations ensuring total mechanical energy conservation. This part of the solution helped us determine how the initial height affected the speed of the block upon reaching the ground.

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Most popular questions from this chapter

You are asked to design a spring that will give a 1160 -kg satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g} .\) The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

A ball is thrown upward with an initial velocity of 15 \(\mathrm{m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

CALC A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta .\) (a) Show that \(U(x)\) can be written as $$U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]$$ Graph \(U(x) .\) Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x)\) , the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x) .\) Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0}\) and \(x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

CALE A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible, (a) Calculate the potential-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0 .\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

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