Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = 
x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s
In short, Your Answer would be Option C
Hope this helps!
A) the resistance is increasing
Hope this helped!
Given the equation for the Speed of a Satellite
v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem
we have:
(square root whole term on right side)
v = G Me
———
r
so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)
v = 7055 m/s (which is reasonable)
so utilize the Kinetic Energy Formula
KE = 1/2mv^2
KE = 1/2(200)(7055)^2
KE = 4.977x10^9 J
