Select the correct answer.

Which equation correctly relates kinetic energy, mass, and velocity?

A. [tex]KE = \frac{1}{2} m^2 v[/tex]
B. [tex]KE = \frac{1}{2} m v^2[/tex]
C. [tex]KE = \frac{1}{2} m v[/tex]
D. [tex]KE = \frac{1}{2} m v^3[/tex]



Answer :

To determine which equation correctly relates kinetic energy, mass, and velocity, let’s consider the physical principles behind kinetic energy.

The kinetic energy ([tex]\( KE \)[/tex]) of a moving object is given by the expression:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]

where:
- [tex]\( KE \)[/tex] is the kinetic energy,
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( v \)[/tex] is the velocity of the object.

Given this relationship, let’s compare it to the options provided in the question.

Option A: [tex]\( KE = \frac{1}{2} m^2 v \)[/tex]

This equation suggests that the kinetic energy is proportional to the square of the mass and linear in velocity. This is incorrect because kinetic energy is not directly proportional to the square of the mass. The relationship must involve the mass and the square of the velocity.

Option B: [tex]\( KE = \frac{1}{2} m v^2 \)[/tex]

This equation matches our known formula for kinetic energy. It correctly relates the mass [tex]\( m \)[/tex] and the square of the velocity [tex]\( v^2 \)[/tex] to kinetic energy. This is the correct answer.

Option C: [tex]\( KE = \frac{1}{2} m v \)[/tex]

This equation suggests that the kinetic energy is directly proportional to both the mass and the velocity. This is incorrect because the velocity term should be squared.

Option D: [tex]\( KE = \frac{1}{2} m v^3 \)[/tex]

This equation implies that kinetic energy is proportional to the mass and the cube of the velocity. This is not correct because the kinetic energy is proportional to the square of the velocity, not the cube.

Based on our analysis, the correct equation that relates kinetic energy, mass, and velocity is provided in Option B:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Thus, the correct answer is:
B. [tex]\( KE = \frac{1}{2} m v^2 \)[/tex]