What role does the entrepreneur play in the market economy
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Answer:
Explanation:
When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing
motion is
Mx''+ βx' + kx = F(t).
Note for an undamped system
β=0,
Then, the differential equation becomes
Mx'' + kx = F(t).
The force is in the form
F=Fo•Sinωo•t
Let solved for the homogeneous or complementary solution, I.e f(t) = 0
Using D operator
MD² + k = 0
MD²=-k
D²=-k/M
Then, D= ±√(-k/m)
D=±√(k/m) •i
So we have a complex root
Therefore, the solution is
x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]
This is simple harmonic motion that once again we prefer to write in the form
x(t) = A•Sin[ √(k/M)t + φ]
Where A=√(C1²+C2²)
and angle φ is defined by the equations
sin φ = C1/A and cos φ = C2/A.
Quantity √(k/M), often denoted by ω, is called the angular frequency.
This is called the natural frequency (ωn) of the system
ωn=√(k/M)
ωn²= k/M
Now, for particular solution
Xp=DSinωo•t
Xp' = Dωo•Cosωo•t
Xp"=-Dωo²•Sinωo•t
Now substituting this into
Mx'' + kx = F(t).
M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t
Now, let solve for D
D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t
D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)
D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)
D=Fo / (-Mωo²+k)
D=Fo / (k-Mωo²)
Divide through by k
D=Fo/k ÷ (1 -Mωo²/k)
Note from above
ωn²= k/M
Therefore,
D=Fo/k ÷ (1-ωo²/ωn²)
D=Fo/k ÷ [1-(ωo/ωn)²]
Then,
Xp=DSinωo•t
Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t
Then the general solution is the sum of the homogeneous solution and particular solution
Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]
Check attachment for the graph of homogeneous, particular and general solution.
Also, check for better way of writing the equations.
Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
Answer:
The distance the block will slide before it stops is 3.3343 m
Explanation:
Given;
mass of bullet, m₁ = 20-g = 0.02 kg
speed of the bullet, u₁ = 400 m/s
mass of block, m₂ = 2-kg
coefficient of kinetic friction, μk = 0.24
Step 1:
Determine the speed of the bullet-block system:
From the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the bullet-block system after collision
(0.02 x 400) + (2 x 0) = v (0.02 + 2)
8 = v (2.02)
v = 8/2.02
v = 3.9604 m/s
Step 2:
Determine the time required for the bullet-block system to stop
Apply the principle of conservation momentum of the system
when the system stops, vf = 0
3.9604 -2.352t = 0
2.352t = 3.9604
t = 3.9604/2.352
t = 1.684 s
Thus, time required for the system to stop is 1.684 s
Finally, determine the distance the block will slide before it stops
From kinematic, distance is the product of speed and time
Now, recall that t = 1.684 s
S = 3.9604(1.684) - 1.176(1.684)²
S = 6.6693 - 3.3350
S = 3.3343 m
Thus, the distance the block will slide before it stops is 3.3343 m