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Nady [450]
2 years ago
11

Write 3 – 4 sentences to describe the kinetic and potential energy of a particle and how the two terms are related.

Physics
1 answer:
mamaluj [8]2 years ago
3 0

Potential energy is position relative, and kinetic energy is motion relative. The primary relationship between the two is their ability to transform into each other. In other words, potential energy transforms into kinetic energy, and kinetic energy converts into potential energy, and then back again.

Potential energy defined as :

The energy stored in a body due to either its position or change in shape is known as potential energy. Water stored in a reservoir is an example of potential energy.

PE=mgh

where,

PE is the potential energy

m is the mass of the body

g is the acceleration due to gravity

h is the height above the surface of the earth

Kinetic energy is defined as :

Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles. Any object in motion is using kinetic energy: a person walking, a thrown baseball, a crumb falling from a table, and a charged particle in an electric field are all examples of kinetic energy at work.

Kinetic Energy Formula:

For the Kinetic formula, Ek, is certainly the energy of a mass, m, motion, of course, is v2.

KE = \frac{1}{2} mv^{2}

KE = Kinetic energy

m = mass of the body

v = velocity of the body

What are the similarities between potential energy and kinetic energy?

One of the most important similarities between potential energy and kinetic energy is that they can both be converted into each other's form of energy. When a force is applied to potential energy it converts to kinetic energy. This means that when an object such as a ball is still it has the potential to move.

Learn more about difference between kinetic and potential energy :

brainly.com/question/23558991

#SPJ4

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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
Mariana [72]

a) F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

R-Fsin \theta - mg = 0 (1)

where

R is the normal reaction of the floor (upward)

F sin \theta is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

F cos \theta - \mu_k R = 0 (2)

where

F cos \theta is the horizontal component of the force F (forward)

\mu_k R is the force of friction (backward)

From (1) we get

R=Fsin \theta +mg

And substituting into (2)

F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

b) \mu_s=cot(\theta)

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

R-Fsin \theta - mg = 0 (1)

F cos \theta - \mu_s R = 0 (2)

In this second case, we want to find the critical value of \mu_s such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, F cos \theta.

Therefore, using the same procedure as before,

R=Fsin \theta +mg

F cos \theta - \mu_s (Fsin \theta +mg) = 0

And solving for \mu_s,

F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}

Now we analyze the expression that we found. We notice that if the force applied F is very large, F sin \theta >> mg, therefore we can rewrite the expression as

\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)

So, this is the critical value of the coefficient of static friction.

8 0
3 years ago
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