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:
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Explanation:
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Answer:
For vector u, x component = 10.558 and y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis
So
Vertical component
So vector u will be u = 10.558 i+12.808 j
Unit vector
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis
So horizontal component
Vertical component
So vector v will be 23.6316 i - 6.464 j
Unit vector of v
Answer:
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