The weight of an object on the moon is one-sixth of its weight on the Earth. The ratio of the kinetic energy of a body on the Earth moving with speed V to that of the same body moving with speed V on the moon is:
The weight of an object on the moon is one-sixth of its weight on the Earth. The ratio of the kinetic energy of a body on the Earth moving with speed V to that of the same body moving with speed V on the moon is
(1/2) m v^2 does not have g in it.
The kinetic energy of a body is given by the formula:
KE = (1/2) mv^2
Where KE is the kinetic energy, m is the mass of the body, and v is its velocity.
Since the weight of the body on the moon is one-sixth of its weight on Earth, we can say that the mass of the body remains the same on both the moon and Earth.
Let's denote the kinetic energy of the body on Earth as KEe and the kinetic energy of the body on the moon as KEm.
Using the formula for kinetic energy, we can express the ratio of the kinetic energies as:
KEm / KEe = [(1/2) m V^2]moon / [(1/2) m V^2]Earth
Since the mass is the same on both Earth and the moon, it cancels out:
KEm / KEe = (1/2) V^2moon / (1/2) V^2Earth
Simplifying further:
KEm / KEe = V^2moon / V^2Earth
Since the question only asks for the ratio, we can conclude that the ratio of the kinetic energy of a body moving with speed V on Earth to that of the same body moving with speed V on the moon is:
V^2moon / V^2Earth
To find the ratio of the kinetic energy of a body on the Earth to that on the moon, we need to understand the relationship between weight and kinetic energy.
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Now, since weight is directly proportional to mass, we can write the equation for weight (W) as:
W = m * g
where g is the acceleration due to gravity.
Given that the weight of an object on the moon is one-sixth of its weight on the Earth, we can write:
W_moon = (1/6) * W_earth
Now, let's assume the mass of the body is the same on both the Earth and the moon. So, m_earth = m_moon = m.
Substituting the weight equations into the kinetic energy equation, we have:
KE_earth = (1/2) * W_earth * v^2 = (1/2) * m * g_earth * v^2
KE_moon = (1/2) * W_moon * v^2 = (1/2) * m * g_moon * v^2
Substituting the weight relationship, we get:
KE_moon = (1/2) * (1/6) * W_earth * v^2
Simplifying, we have:
KE_moon = (1/12) * W_earth * v^2
Finally, we can find the ratio of the kinetic energy of the body on the Earth to that on the moon by dividing the two equations:
KE_earth / KE_moon = [(1/2) * m * g_earth * v^2] / [(1/12) * W_earth * v^2]
The mass, velocity, and gravitational acceleration cancel out, so we're left with:
KE_earth / KE_moon = (1/2) / (1/12) = 6
Therefore, the ratio of the kinetic energy of a body on the Earth moving with speed V to that of the same body moving with speed V on the moon is 6:1.