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A \(0.60-\mathrm{kg}\) particle has a speed of \(2,0 \mathrm{~m} / \mathrm{s}\) at point \(A\) and a kinetic energy of \(7.5 \mathrm{~J}\) at point \(\mathrm{B}\). 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
(a) The kinetic energy at point A is 1.2 J. (b) The speed at point B is 5 ms^{-1}. (c) The total work done on the particle as it moves from A to B is 6.3 J.

Step by step solution

01

Determining the kinetic energy at point A

The kinetic energy at point A can be calculated using the kinetic energy formula \(K.E. = 0.5mv^2\). Here the mass \(m = 0.60 kg\) and the speed \(v = 2.0 ms^{-1}\). Therefore, substituting these values into the formula, the kinetic energy at A is \(K.E. = 0.5* 0.60 kg *(2.0 ms^{-1})^2 = 1.2 J\).
02

Determining the speed at point B

To determine the speed at point B, we need to rearrange the kinetic energy formula to make \(v\) the subject. The rearranged formula is \(v = \sqrt{2K.E./m}\). Substituting the given kinetic energy at B (\(7.5 J\)) and the mass (\(0.60 kg\)) into this formula gives \(v = \sqrt{2*7.5 J/0.60 kg} = 5 ms^{-1}\).
03

Determining the total work done on the particle

The total work done on the particle as it moves from A to B is the change in kinetic energy (kinetic energy at B - kinetic energy at A). Therefore, the work done is \(7.5 J - 1.2 J = 6.3 J\).

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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 of physics that explains the relationship between work done on an object and the change in its kinetic energy. When a force acts on an object to move it, work is done, and as a result, the object's kinetic energy changes.

The work done (typically represented as \( W \)) on an object is equal to the change in its kinetic energy (\( \Delta KE \)). Thus, the theorem can be expressed as:
  • \( W = \Delta KE = KE_{final} - KE_{initial} \)
In the exercise, the work done on the particle from point A to B can be calculated as the difference in kinetic energy between those points, which is \( 7.5\, J - 1.2\, J = 6.3\, J \).

This theorem highlights that whenever there is work done on an object, there is a corresponding change in kinetic energy. This is crucial for understanding how forces and energy relate in dynamic systems.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It focuses on properties such as velocity, acceleration, and displacement. It tells us "how" an object moves, but not "why."

In the exercise, we are interested in how the particle's speed changes between points A and B:
  • At point A, the speed was given as \( 2.0\, m/s \), and we used this with its mass to calculate the initial kinetic energy.
  • At point B, knowing the kinetic energy \( (7.5\, J) \), we rearranged the kinetic energy formula to find the speed \( v \).
The speed of the particle at B was calculated to be \( 5\, m/s \).

This process involves using the kinematic equation derived from kinetic energy to establish a link between velocity and kinetic concepts. Linear Kinematics serves as a foundational building block for more complex topics like dynamics and provides crucial insights into the behavior of objects in motion.
Particle Dynamics
Particle Dynamics involves analyzing forces acting on particles, their resulting motion, and energy changes due to these forces. It's essential for understanding how and why objects move the way they do under various forces.

In this exercise, we addressed the changes in the particle's kinetic energy as an effect of work done. The change in kinetic energy between points A and B (6.3 J) represents the total work done by external forces on the particle.
  • This reflects an interplay of forces that caused the particle to accelerate and increase its speed from 2.0 m/s to 5.0 m/s.
  • Understanding dynamics allows one to predict future motion and analyze complex systems.
By evaluating the forces and the work done, we understand particle dynamics, which is central to fields like engineering and physics. It informs how objects behave under various conditions and interacts with forces, crucial for problem-solving in applied physics.

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Most popular questions from this chapter

(a) A block with a mass \(m\) is pulled along a horizontal surface for a distance \(x\) by a constant force \(\overrightarrow{\mathbf{F}}\) at an angle \(\theta\) with respect to the horizontal. The coefficient of kinetic friction between block and table is \(\mu_{k} .\) Is the force exerted by friction equal to \(\mu_{h} m g_{2}^{2}\) If not, what is the force exerted by friction? (b) How much work is done by the friction force and by \(\vec{F}\) ? (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let. \(m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}, F=15.0 \mathrm{~N}\), and \(\mu_{k}=\) \(0.400\), and find the answers to parts (a) and (b).

A \(70-\mathrm{kg}\) base runner begins his slide into second base when he is moving at a speed of \(4,0 \mathrm{~m} / \mathrm{s}\). The cocfficient of friction between his clothes and Farth is \(0.70 .\) He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? (b) How far does he slide?

A 70 -kg diver steps off a \(10-\mathrm{m}\) tower and drops from rest straight down into the water. If he comes to rest \(5.0 \mathrm{~m}\) beneath the surface, determine the average resistive force exerted on him by the water.

A weight lifter lifts a \(350-\mathrm{N}\) set of weights from ground level to a position over his head, a vertical distance of \(2.00 \mathrm{~m}\). How much work does the weight lifter do, assuming he moves the weights at constant speed?

Find the height from which you would have to drop a ball so that it would have a speed of \(9.0 \mathrm{~m} / \mathrm{s}\) just before it hits the ground.

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