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3 years ago

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A single conservative force acts on a 4.50-kg particle within a system due to its interaction with the rest of the system. The e

quation Fx = 2x + 4 describes the force, whereFx is in newtons and x is in meters. As the particle moves along the x axis from x = 0.90 m to x = 5.15 m, calculate the following.
(a) the work done by this force on the particle? (J)

(b) the change in the potential energy of the system? (J)

(c) the kinetic energy the particle has at x = 5.15 m if its speed is 3.00 m/s at x = 0.90 m? (J)

1 answer:

den301095 [7]3 years ago

8 0

Answer: (a) The work done by this force on the particle is 42.71 J.

(b) The change in the potential energy of the system is -42.71 J.

(c) The kinetic energy the particle is 62.96 J.

Explanation:

(a) For the given situation, expression for work done is as follows.

W = $\int_{0.9}^{5.15}Fdx$

= $\int_{0.9}^{5.15}(2x + 4)dx$

= $[2\frac{x^{2}}{2} + 4x]^{5.15}_{0.9}$

= $(x^{2} + 4x)^{5.15}_{0.9}$

= $[(5.15)^{2} + 4(5.15) - (0.9)^{2} - 4(0.9)]$

= 26.52 + 20.6 - 0.81 - 3.6

= 42.71 J

Hence, the work done by this force on the particle is 42.71 J.

(b) Expression for a conservative force is as follows.

F = $-\frac{dU}{dx}$

dU = -Fdx

$\int_{0.9}^{5.15}dU$ = $\int_{0.9}^{5.15}Fdx$

$\int_{0.9}^{5.15}dU$ = -42.71 J

Therefore, the change in the potential energy of the system is -42.71 J.

(c) According to the work energy theorem,

W = $\Delta K.E$

$K.E_{0.9} - K.E_{5.15}$ = W

$K.E_{0.9} = W + K.E_{5.15}$

= $42.71 + \frac{1}{2}mu^{2}$

where, u = velocity of the mass at x = 0.9 m

u = 3.0 m/s, m = 4.50 kg

As, $K.E_{0.9} = W + K.E_{5.15}$

= $42.71 + \frac{1}{2} \times 4.50 \times (3)^{2}$

= 62.96 J

Therefore, the kinetic energy the particle is 62.96 J.

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In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

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where;

$\mu$ is the dipole moment
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Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
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3. Promises rapid weight loss in a short period of time.

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3. More accumulation of fats in the body in the long run after a short term normal BMI (Body mass index).

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A brick of mass 5 kg is released from rest at a height of 3 m. How fast is it going when it hits the ground? Acceleration due to

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Taking into account the definition of kinetic, potencial and mechanical energy, when the brick hits the ground, it has a speed of 7,668 m/s.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

Ec is the kinetic energy, which is measured in Joules (J).
m is the mass measured in kilograms (kg).
v is the speed measured in meters over seconds (m/s).

<h3>Potential energy</h3>

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

Ep is the potential energy in joules (J).
m is the mass in kilograms (kg).
h is the height in meters (m).
g is the acceleration of fall in m/s².

<h3>Mechanical energy</h3>

Finally, mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

<h3>Principle of conservation of mechanical energy </h3>

The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative (a force is conservative when the work it does on a body depends only on the initial and final points and not the path taken to get from one to the other.)

Therefore, if the potential energy decreases, the kinetic energy will increase. In the same way, if the kinetics decreases, the potential energy will increase.

<h3>This case</h3>

A brick of mass 5 kg is released from rest at a height of 3 m. Then, at this height, the brick of mass has no speed, so the kinetic energy has a value of zero because it depends on the speed or moving bodies. But the potential energy is calculated as:

Ep= 5 kg× 9.8 $\frac{m}{s^{2} }$ × 3 m

Solving:

<u><em>Ep= 147 J</em></u>

So, the mechanical energy is calculated as:

Potential energy + kinetic energy = total mechanical energy

147 J + 0 J= total mechanical energy

147 J= total mechanical energy

The principle of conservation of mechanical energy can be applied in this case. Then, when the brick hits the ground, the mechanical energy is 147 J. In this case, considering that the height is 0 m, the potential energy is zero because this energy depends on the relative height of the object. But the object has speed, so it will have kinetic energy. Then:

Potential energy + kinetic energy = total mechanical energy

0 J + kinetic energy= 147 J

kinetic energy= 147 J

Considering the definition of kinetic energy:

½ 5 kg×v²= 147 J

$v=\sqrt{\frac{2x147 J}{5 kg} }$

v=7.668 m/s

Finally, when the brick hits the ground, it has a speed of 7,668 m/s.

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