/*! 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 16 A \(0.60-\mathrm{kg}\) particle ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(0.60-\mathrm{kg}\) particle has a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) at point \(A\) and a kinetic energy of 7.5 J at point \(B\). What is (a) its kinetic energy at \(A\) ? (b) Its speed at point \(B\) ? (c) The total work done on the particle as it moves from \(A\) to \(B\) ?

Short Answer

Expert verified
The kinetic energy at point A is 1.2 Joules. The speed at point B is 4.0 m/s. The total work done on the particle as it moves from A to B is 6.3 Joules.

Step by step solution

01

Calculate kinetic energy at A

Use the formula for the kinetic energy, \(KE = \frac{1}{2} mv^2\) where m is mass, 0.60 kg and v is speed, 2.0 m/s. So, \(KE_{A} = \frac{1}{2} * 0.60 kg * (2.0 m/s)^2\)
02

Calculate speed at B

To calculate speed at B, rearrange the kinetic energy formula, so \(v_{B} = \sqrt{ \frac{2*KE_{B}}{m} }\). With KE at B equals 7.5 J and mass=0.60 kg, we substitute these into the rearranged formula.
03

Calculate work done

The total work done on the particle as it moves from A to B equals to the change in Kinetic Energy, so \(Work = KE_{B} - KE_{A}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 physics that describes the relationship between the work done on an object and the change in its kinetic energy. In simpler terms, it tells us that the work done by the forces acting on an object results in a change in the object’s motion or kinetic energy. For example, when you apply a force to a ball and move it, you have done work on the ball, increasing its kinetic energy and making it move faster.

The mathematical expression of the work-energy principle is given by the equation:
  • Work Done (W) = Change in Kinetic Energy (ΔKE)
where the change in kinetic energy is calculated as the final kinetic energy (KE_{final}) minus the initial kinetic energy (KE_{initial}). In the context of the original exercise, the total work done on the particle as it moves from point A to point B results in an increase in its kinetic energy from the value at A to the value at B.

To express it using equations, we have:
Work = KE_{B} - KE_{A}where KE_{B} is the kinetic energy at point B and KE_{A} is the initial kinetic energy at point A. This equation is directly used to answer part (c) of the exercise, where the kinetic energy at A (KE_{A}) is found using the kinetic energy formula and then subtracted from the given kinetic energy at B (KE_{B}) to find the work done.
Conservation of Energy
The principle of conservation of energy states that energy can neither be created nor destroyed, only transformed from one form to another. In a closed system, where no energy enters or leaves, the total energy remains constant. This is a key concept in understanding a wide array of physical systems and plays a crucial role in solving problems involving forces and motion. When we apply this principle to kinetic energy, we can understand how potential energy can be converted into kinetic energy and vice versa, without any loss of total energy.

For instance, if a particle is only under the influence of conservative forces, such as gravity, the mechanical energy (sum of potential energy and kinetic energy) of the system remains constant. In the case of our exercise, if we consider that no other energy transfer (like heat or sound) is involved, the work done on the particle can be viewed as the conversion of energy from one form into kinetic energy. This is important for imagining the trajectory of the particle from point A to point B, assuming no energy is lost, but rather, it is 'conserved' and transformed into motion.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause this motion. It involves the study of position, velocity, acceleration, and time, and the relationships between these quantities. Knowing the kinematic equations allows us to predict the future position of moving objects if the initial conditions and the motion characteristics are known.

In solving our exercise, kinematics came into play when determining the speed of the particle at point B. We had to consider the kinematic relationship that links speed with kinetic energy. By rearranging the kinetic energy formula, we derived the speed in terms of kinetic energy and mass, giving us an understanding of how the speed changes with the kinetic energy of the particle. This direct relationship between speed and kinetic energy is crucial for solving part (b) of the exercise:
v_{B} = sqrt{ frac{2*KE_{B}}{m} }. Hence, kinematics principles help us deduce the final speed based on the initial conditions and the applied work, illustrating its importance in analyzing motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

BIO The chin-up is one exercise that can be used to strengthen the biceps muscle. This muscle can exert a force of approximately \(800 \mathrm{~N}\) as it contracts a distance of \(7.5 \mathrm{~cm}\) in a \(75-\mathrm{kg}\) male. \({ }^{3}\) How much work can the biceps muscles (one in cach arm) perform in a single contraction? Compare this amount of work with the energy required to lift a \(75-\mathrm{kg}\) person \(40 \mathrm{~cm}\) in performing a chin-up. Do you think the biceps muscle is the only muscle involved in performing a chin- up?

A \(50-\mathrm{kg}\) pole vaulter running at \(10 \mathrm{~m} / \mathrm{s}\) vaults over the bar. Her speed when she is above the bar is \(1.0 \mathrm{~m} / \mathrm{s}\). Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

GP A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m}\). A block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure \(5.20\). The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (c) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway. He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point \(10.0 \mathrm{~m}\) above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a \(5.00-\mathrm{m}\) length of the cord, the jumper finds that his body weight stretches it by \(1.50 \mathrm{~m}\). He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?

A \(1.50 \times 10^{3}-\mathrm{kg}\) car starts from rest and accelerates uniformly to \(18.0 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\). Assume that air resistance remains constant at \(400 \mathrm{~N}\) during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at \(t=12.0 \mathrm{~s}\), just before the car stops accelerating.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.