You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).
And you know that the potential energy, PE, is [ 1/2 ] k (x^2)
Then, use x = A, to calculate the PE in the point where ME = PE.
ME = PE = [1/2] k (A)^2.
At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2
=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME
So, if PE is 1/4 of ME, KE is 3/4 of ME.
And the answer is 3/4
Answer:
A. 16.9 m
Explanation:
I think this is the answer i am not sure
but hope it helps
Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
