Equation For Kinetic Energy

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The work required to accelerate a body of a specified mass from rest to its recent velocity is known as kinetic energy. The kinetic energy of an object is the extra energy which the object possesses due to its action. The kinetic energy of one object is entirely frame-dependent.

The kinetic energy of a rigid body with mass and speed v is given by mv2 / 2, where the v is less than the speed of light.
Equation for Kinetic Energy:

Kinetic energy of rigid bodies:

The kinetic energy of a non-rotating rigid body or an object is written as the following equation,

Ek = (1/2) mv2

Where m is the mass is m, v is the speed or velocity of the rigid body and Ek is called as the kinetic energy.

Derivation:

The work done by accelerating an object at the time interval dt is given by the vector product of force and displacement.

F.dx = F.vdt = (dp/dt).vdt = v.dp = v.d (mv)

When we apply the product rule, then we get,

d(v.v) = (dv).v + v.(dv) = 2(v.dv)

v.d(mv) = (m/2)d(v.v) = (m/2)dv2 = d(mv2 /2).

The above equation is total differential and when we integrate, ...
... we get

Ek = ? F.dx = ? v.dp = mv2 / 2.
Examples for Kinetic Energy:

Example:

Calculate the kinetic energy of a rigid body with a mass of 100 kg which is travelling at 20 m/sec.

Solution:

Kinetic energy Ek = (1/2) mv2

Mass = 100 kg and speed = 20 m/sec

= (1/2)(100)(202)

= 20000.

The kinetic energy is expressed in joules

Kinetic energy Ek = 20000 joules.

The square of the speed of the object increases the kinetic energy.

The kinetic energy of an object is associated to its momentum by the equation

Ek = p2 / 2m

Where the mass of the body is m and the momentum is p.

Rotations in systems:

The addition of the body’s centre of mass translational kinetic energy and the energy of rotation gives the total kinetic energy.The equation of total kinetic energy is written as follows.

Ek = Et + Er

Where,

The rotational energy is Er

The translational kinetic energy is Et

The total kinetic energy is Ek.

The standard form of the quadratic equation model is discussed below. In mathematics, a quadratic equation is one of a polynomial with the degree two. The standard model equation form is given as

ax2+bx+c=0

Where x from the above equation represent a variable, and a, b, and c, are the constants, with a ? 0. (If a = 0, the equation becomes a linear equation.). The constants a, b, and c, are known as, the quadratic coefficient, the linear integer and the constant term or free term. The word "quadratic" is a Latin word comes from quadratus; means "square." Quadratic equations may be calculated by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.

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Example Problems on Standard Model Equation:

Example 1:

Solve the standard equation model.

Sqrt (a + 1) = 4

Solution:

Given equation is
sqrt (a + 1) = 4

Squaring on both sides, then the above equation becomes
[sqrt (a + 1)] 2 = 4 2

Simplify the above equation
a + 1 = 16

Solve for a.
a = 15

To check the left hand side (LHS) of the given equation when a = 15

LS = sqrt (a + 1) = sqrt (15 + 1) = 4

to check the right hand side (RHS) of the given equation when a = 15

RHS = 4

When a = 15, the left and right sides of the given equation are equal, so a = 15 is a solution for the given equation.

Example 2:

Solve the standard equation model.

Sqrt (3 z + 1) = z - 3

Solution:

Given equation is
sqrt ( 3 z + 1) = z - 3

Squaring on both sides, then the above equation becomes
[sqrt ( 3 z + 1) ] 2 = (z - 3) 2

Simplify the above equation
3 z + 1 = z 2 - 6 z + 9

Rewrite the above equation in factor form.
z 2 - 9 z + 8 = 0

The above form is a quadratic equation with 2 solutions
z = 8 and z = 1
Practice Problems on Standard Model Equation:

1) Solve the standard equation model.

Sqrt (2 y + 15) = 5
Answer: y = 5

2) Solve the standard equation model.

Sqrt (4 y - 3) = x – 2

Answer: y = 7

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