Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:
c
Explanation:
without force motion won't take place
Answer:
20.2 seconds
Explanation:
The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.
The crate is in free fall, so a = -9.8 m/s².
The crate falls downward, so Δx = -2000 m.
Find: t, the time it takes for the crate to land.
Δx = v₀ t + ½ at²
-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 20.2 s
It takes 20.2 seconds for the crate to land.
