What is the initial height big an object if, when dropped from rest , its kinetic energy is 21J 2/3 of the way towards the ground? The mass is 5
mg(2/3h)=21
solve for h, m, g are given.
dropped from h, total energy = m g h
at h/3 total energy = m g h = m g h/3 + 21
(2/3) m g h = 21
(2/3)(5)(9.8) h = 21
solve for h
To solve this problem, we need to use the concept of potential energy and kinetic energy. The equation for potential energy is given by:
Potential Energy (PE) = mass (m) * gravity (g) * height (h)
The equation for kinetic energy is given by:
Kinetic Energy (KE) = (1/2) * mass (m) * velocity^2
Given that the kinetic energy of the object is 21J 2/3 of the way towards the ground, we can set up the equation as follows:
KE = 21J * (2/3)
Now, we can rearrange the equation for kinetic energy and solve for the velocity:
21J * (2/3) = (1/2) * 5kg * velocity^2
Simplifying the equation:
14J = 2.5kg * velocity^2
Divide both sides by 2.5kg:
velocity^2 = 14J / 2.5kg
velocity^2 = 5.6m/s^2
Now, we can solve for the final velocity. Since the object is dropped from rest, the initial velocity (u) is 0 m/s. The final velocity (v) can be calculated using the equation:
v^2 = u^2 + 2 * acceleration * distance
Since the initial velocity is 0, the equation simplifies to:
v^2 = 2 * acceleration * distance
Plugging in the values:
5.6m/s^2 = 2 * 9.8m/s^2 * distance
Divide both sides by 2 * 9.8m/s^2:
distance = 5.6m/s^2 / (2 * 9.8m/s^2)
distance = 0.286 meters
So, the distance from the initial height to the point with 2/3 of the kinetic energy is 0.286 meters.