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. V = 2 m/s
Explanation:
Using the conservation of energy:
so:
Mgh = 
where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.
Also we know that:
V = WR
Where R is the radius of the disk, so:
W = V/R
Also, the moment of inertia of the disk is equal to:
I = 
I = 
I = 10 kg*m^2
so, we can write the initial equation as:
Mgh = 
Replacing the data:
(5kg)(9.8)(0.3m) = 
solving for V:
(5kg)(9.8)(0.3m) = 
V = 2 m/s
Answer:
The speed of the 11.5kg block after the collision is V≅4.1 m/s
Explanation:
ma= 4.8 kg
va1= 7.3 m/s
va2= - 2.5 m/s
mb= 11.5 kg
vb1= 0 m/s
vb2= ?
vb2= ( ma*va1 - ma*va2) / mb
vb2= 4.09 m/s ≅ 4.1 m/s
You would want it to be greater than D. friction force
It needs be greater than the friction applied to it.
Answer:
Tension is also known as Force...
and Force is mass× acceleration.
so....1000×0.70=700N
