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Chapter 7: Problem 24

A 0.600 -kg particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) at point (A) and kinetic energy of \(7.50 \mathrm{J}\) at point (B). What is (a) its kinetic energy at (A) ? (b) its speed at (B)? (c) the total work done on the particle as it moves from (A) to (B)?

Short Answer

Expert verified
The kinetic energy at (A) is 1.20 J, the speed at (B) is 5.0 m/s, and the total work done on the particle as it moves from (A) to (B) is 6.30 J.

Step by step solution

01

Calculate Kinetic Energy at (A)

First, use the formula for kinetic energy to find the energy at point (A). The formula is \( K.E. = \frac{1}{2}mv^2 \). Substituting given values, \( K.E._A = \frac{1}{2} * 0.600kg * (2.00ms^-1)^2 = 1.2J \)
02

Compute the Speed at (B)

To find the speed at point (B), re-arrange the kinetic energy formula to solve for velocity \( v = \sqrt{\frac{2K.E.}{m}} \). Substituting the given values, \( v_B = \sqrt{\frac{2*7.5J}{0.600kg}} = 5.0ms^-1 \)
03

Find the Total Work Done

This can be determined using the work-energy theorem: the work done \(W\) is equal to the change in kinetic energy \(\Delta K.E.\), i.e., \( W = \Delta K.E. = K.E._B - K.E._A \). By substituting the calculated kinetic energy values, the total work done on the particle is \( W = 7.50J - 1.20J = 6.30J \).

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Key Concepts

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

Work-Energy Theorem
The work-energy theorem is a fundamental principle in physics that ties together the concept of work and kinetic energy. According to this theorem, the work done on an object is equal to the change in its kinetic energy. This relationship can be expressed mathematically as: \[ W = \Delta K.E. = K.E._\text{final} - K.E._\text{initial} \]Where:
  • \( W \) is the work done on the object.
  • \( K.E._\text{final} \) is the final kinetic energy.
  • \( K.E._\text{initial} \) is the initial kinetic energy.

In the context of our exercise, the particle is moving from point A to point B. The work done on the particle equals the change in its kinetic energy from its state at point A to point B. Understanding this concept is crucial as it helps in solving problems related to energy conversions and mechanical work in physics.
Kinetic Energy Formula
Kinetic energy is the energy that an object possesses due to its motion. The formula to calculate kinetic energy is quite simple and elegant:\[ K.E. = \frac{1}{2}mv^2 \]Here:
  • \( K.E. \) stands for kinetic energy.
  • \( m \) is the mass of the object.
  • \( v \) is the velocity of the object.

Let's apply this formula to point A in the problem. By substituting the given values, the kinetic energy at point A is found to be 1.2 Joules. Knowing the kinetic energy at a particular point allows us to rearrange the formula to solve for unknown variables, such as velocity or further energy states. The exercise also makes use of this formula to find the speed at point B once the kinetic energy at B is known.
Work Done
Work done on an object is a measure of energy transfer. It occurs when a force causes an object to move. Conceptually, work done can be calculated when we know the change in an object's kinetic energy. Applying this to our problem, we use the formula derived from the work-energy theorem:\[ W = K.E._B - K.E._A \]Where:
  • \( W \) is the work done on the particle.
  • \( K.E._B \) is the kinetic energy at point B.
  • \( K.E._A \) is the kinetic energy at point A.

When calculating the work done on the particle traveling from point A to B, we take the difference between the kinetic energies at these two points. The result, 6.30 Joules, indicates the energy transferred to the particle through work as it moves. This shows a practical application of how the work-energy theorem helps us understand energy transformations in motion-based problems.

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Most popular questions from this chapter

The direction of any vector \(\mathbf{A}\) in three-dimensional space can be specified by giving the angles \(\alpha, \beta,\) and \(\gamma\) that the vector makes with the \(x, y,\) and \(z\) axes, respectively. If \(\mathbf{A}=\) \(A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}+A_{z} \hat{\mathbf{k}},\) (a) find expressions for \(\cos \alpha, \cos \beta,\) and cos \(\gamma\) (these are known as direction cosines), and (b) show that these angles satisfy the relation \(\cos ^{2} \alpha+\cos ^{2} \beta+\) \(\cos ^{2} \gamma_{n}=1 .\) (Hint: Take the scalar product of A with \(\hat{\mathbf{i}}, \hat{\mathbf{j}}\) and k separately.)

A bead at the bottom of a bowl is one example of an object in a stable equilibrium position. When a physical system is displaced by an amount \(x\) from stable equilibrium, a restoring force acts on it, tending to return the system to its equilibrium configuration. The magnitude of the restoring force can be a complicated function of \(x .\) For example, when an ion in a crystal is displaced from its lattice site, the restoring force may not be a simple function of \(x .\) In such cases we can generally imagine the function \(F(x)\) to be expressed as a power series in \(x,\) as \(F(x)=-\left(k_{1} x+k_{2} x^{2}+k_{3} x^{3}+\ldots\right) .\) The first term here is just Hooke's law, which describes the force exerted by a simple spring for small displacements. For small excursions from equilibrium we generally neglect the higher order terms, but in some cases it may be desirable to keep the second term as well. If we model the restoring force as \(F=-\left(k_{1} x+k_{2} x^{2}\right),\) how much work is done in displacing the system from \(x=0\) to \(x=x_{\max }\) by an applied force \(-F ?\)

More than 2300 years ago the Greek teacher Aristotle wrote the first book called Physics. Put into more precise terminology, this passage is from the end of its Section Eta: Let \(\mathscr{P}\) be the power of an agent causing motion; \(w\), the thing moved; \(d\), the distance covered; and \(\Delta t\), the time interval required. Then (1) a power equal to \(\mathscr{P}\) will in a period of time equal to \(\Delta t\) move \(w / 2\) a distance \(2 d ;\) or (2) it will move \(w / 2\) the given distance \(d\) in the time interval \(\Delta t / 2 .\) Also, if (3) the given power \(\mathscr{P}\) moves the given object \(w\) a distance \(d / 2\) in time interval \(\Delta t / 2,\) then \((4) \mathscr{P} / 2\) will move \(w / 2\) the given distance \(d\) in the given time interval \(\Delta t.\) (a) Show that Aristotle's proportions are included in the equation \(\mathscr{P} \Delta t=\) bwd where \(b\) is a proportionality constant. (b) Show that our theory of motion includes this part of Aristotle's theory as one special case. In particular, describe a situation in which it is true, derive the equation representing Aristotle's proportions, and determine the proportionality constant.

A \(15.0\) -kg block is dragged over a rough, horizontal surface by a \(70.0\) - \(\mathrm{N}\) force acting at \(20.0^{\circ}\) above the horizontal. The block is displaced \(5.00 \mathrm{~m}\), and the coefficient of kinetic friction is \(0.300\). Find the work done on the block by (a) the 70 - \(\mathrm{N}\) force, (b) the normal force, and (c) the gravitational force. (d) What is the increase in internal energy of the block- surface system due to friction? (e) Find the total change in the block's kinetic energy.

A \(700-\mathrm{N}\) Marine in basic training climbs a \(10.0-\mathrm{m}\) vertical rope at a constant speed in 8.00 s. What is his power output?
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