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Just before striking the ground, a \(2.00-\mathrm{kg}\) mass has \(400 \mathrm{~J}\) of \(\mathrm{KE}\). If friction can be ignored, from what height was it dropped?

Short Answer

Expert verified
The initial height was 20.41 meters.

Step by step solution

01

Understand Kinetic Energy Formula

The Kinetic Energy (KE) of an object is given by the formula: \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object just before it hits the ground. However, in this problem, we are given the kinetic energy and need to find the initial height when it was dropped.
02

Understand Energy Conservation

Since friction is ignored, the principle of conservation of energy applies. This means that the potential energy (PE) at the starting height is equal to the kinetic energy (KE) just before it strikes the ground. So, \( PE = KE = 400 \, \mathrm{J} \).
03

Potential Energy Formula

The potential energy at a height \( h \) is given by: \( PE = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (\(9.8 \, \mathrm{m/s^2}\)), and \( h \) is the height from which it was dropped.
04

Set Up the Equation

Set the potential energy equal to the given kinetic energy: \( mgh = 400 \, \mathrm{J} \). Substitute the known values: \( 2.00 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times h = 400 \, \mathrm{J} \).
05

Solve for Height

Rearrange the equation to solve for \( h \): \[ h = \frac{400}{2.00 \times 9.8} \].Calculate \( h \): \( h = \frac{400}{19.6} = 20.41 \, \mathrm{m} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. It's a fundamental concept in physics that helps us understand how energy transforms from one form to another through movement. The formula for kinetic energy is:
  • \( KE = \frac{1}{2} m v^2 \)
In this formula, \( m \) signifies the mass of the object, and \( v \) denotes its velocity. The kinetic energy calculation is crucial in problems involving moving objects, such as a falling mass just before it hits the ground.
When an object hits the surface, all its potential energy is converted into kinetic energy, which is why we use this value to find out more about the object's journey prior to the impact.
Being able to understand and calculate kinetic energy allows us to further explore how energy behaves when objects move, fall, or collide.
Potential Energy
Potential energy is the stored energy in an object due to its position or configuration. In the context of an object being at a certain height above the ground, this is often referred to as gravitational potential energy. The formula used here is:
  • \( PE = mgh \)
In this equation, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (\(9.8 \, \mathrm{m/s^2}\) on Earth), and \( h \) is the height relative to a reference point. Potential energy is relevant in physics problems because it represents the capacity for an object to do work due to its position. When an object falls, this energy is converted to kinetic energy, demonstrating the principle of conservation of energy.
Understanding potential energy is vital as it lays the groundwork for solving problems related to energy transformation and conservation.
Physics Problem Solving
Physics problem-solving often involves understanding and applying basic principles to find the solution to more complex scenarios. When tackling physics exercises, like the one given, you should:Start by identifying the known quantities and what you're solving for. In the example, you know the mass and kinetic energy, and you need to find the initial height of a falling object.
  • Utilize the conservation of energy principle which tells us that in the absence of friction, energy is conserved and can transform from potential to kinetic.
  • Apply the appropriate formulas for kinetic and potential energy to set up equations that relate them.
  • Finally, rearrange the equations to solve for the unknown variable. In this case, solving for height \( h \).
Physics problem-solving not only involves mathematical calculations but also requires reasoning through the physical scenario to understand how different factors interact. This process strengthens critical thinking and reinforces an understanding of fundamental physics concepts.

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Most popular questions from this chapter

Figure \(6-3\) shows a bead sliding on a wire. If friction forces are negligible and the bead has a speed of \(200 \mathrm{~cm} / \mathrm{s}\) at \(A\), what will be its speed \((a)\) at point \(B ?\) (b) At point \(C\) ? The energy of the bead is conserved, so we can write Change in \(\mathrm{KE}+\) change in \(\mathrm{PE}_{\mathrm{G}}=0\) $$ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}+m g\left(h_{f}-h_{i}\right)=0 $$ (a) Here, \(v_{i}=2.0 \mathrm{~m} / \mathrm{s}, h_{i}=0.80 \mathrm{~m}\), and \(h_{f}=0\). Using these values, while noticing that \(m\) cancels out, gives \(v_{f}=4.4 \mathrm{~m} / \mathrm{s}\) (b) Here, \(v_{b}=2.0 \mathrm{~m} / \mathrm{s}, h_{i}=0.80 \mathrm{~m}\), and \(h_{f}=0.50 \mathrm{~m} .\) Using these values leads to \(v_{f}=3.1 \mathrm{~m} / \mathrm{s}\). Alternative Method Since energy, \(\mathrm{E}\), is conserved, $$ \begin{aligned} \mathrm{E}_{i} &=\mathrm{E}_{f} \\ \mathrm{KE}_{i}+\mathrm{PE}_{\mathrm{G}} &=\mathrm{KE}_{f}+\mathrm{PE}_{\mathrm{G} f} \\ \frac{1}{2} m v_{i}^{2}+m g h_{i} &=\frac{1}{2} m v_{f}^{2}+m g h_{f} \\ \frac{1}{2} v_{1}^{2}+g h_{i} &=\frac{1}{2} v_{f}^{2}+g h_{f} \\ (2.00 \mathrm{~m} / \mathrm{s})^{2}+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(0.80 \mathrm{~m}-h_{f}\right)=v_{f}^{2} \end{aligned} $$

A \(10.0\) -kg block is launched up a \(30.0^{\circ}\) inclined plane at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). As it slides it loses \(200 \mathrm{~J}\) to friction. How far along the incline will it travel before coming to rest?

How much work is done against gravity in lifting a \(3.0-\mathrm{kg}\) object through a vertical distance of \(40 \mathrm{~cm}\) ? An external force is needed to lift an object. If the object is raised at constant speed, the lifting force must equal the weight of the object. The work done by the lifting force is referred to as work done against gravity. Because the lifting force is \(m g\), where \(m\) is the mass of the object, $$ \text { Work }=(m g)(h)(\cos \theta)=(3.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(0.40 \mathrm{~m})(1)=12 \mathrm{~J} $$ In general, the work done against gravity in lifting an object of mass \(m\) through a vertical distance \(h\) is \(m g h\).

How large a force is required to accelerate a \(1300-\mathrm{kg}\) car from rest to a speed of \(20 \mathrm{~m} / \mathrm{s}\) in a horizontal distance of \(80 \mathrm{~m}\) ?

A \(0.50-\mathrm{kg}\) block slides across a tabletop with an initial velocity of \(20 \mathrm{~cm} / \mathrm{s}\) and comes to rest in a distance of \(70 \mathrm{~cm}\). Find the average friction force that retarded its motion. The \(\mathrm{KE}\) of the block is decreased because of the slowing action of the friction force. That is, Change in \(\mathrm{KE}\) of block \(=\) Work done on block by friction force $$ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}=F_{\mathrm{f}} s \cos \theta $$ Because the friction force on the block is opposite in direction to the displacement, \(\cos \theta=-1\). Using \(v_{f}=0\), \(v_{i}=0.20 \mathrm{~m} / \mathrm{s}\), and \(s=0.70 \mathrm{~m}\), the above equation becomes $$ 0-\frac{1}{2}(0.50 \mathrm{~kg})(0.20 \mathrm{~m} / \mathrm{s})^{2}=\left(F_{\mathrm{f}}\right)(0.70 \mathrm{~m})(-1) $$ from which \(F_{t}=0.014 \mathrm{~N}\).

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