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:
Explanation:
Given
mas of car=870 kg
coffee mug mass=0.47 kg
coefficient of static friction between mug and roof 
Coefficient of kinetic Friction 
maximum car acceleration is 
here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

Answer:
a = -1 m/s^2
Explanation:
Vi = 75 m/s
Vf = 25 m/s
t = 50 s
Plug those values into the following equation:
Vf = Vi + at
25 = 75 + 50a
---> a = -1 m/s^2
The minimum value of the coefficient of static friction between the block and the slope is 0.53.
<h3>Minimum coefficient of static friction</h3>
Apply Newton's second law of motion;
F - μFs = 0
μFs = F
where;
- μ is coefficient of static friction
- Fs is frictional force
- F is applied force
μ = F/Fs
μ = F/(mgcosθ)
μ = (250)/(50 x 9.8 x cos15)
μ = 0.53
Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.
Learn more about coefficient of friction here: brainly.com/question/20241845
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