Answer:2987.7 J
Explanation:
Given
sled weight is
Tension in rope
Time interval
velocity of sled
here speed is equal to velocity which is constant therefore acceleration is zero
thus
Distance traveled in
considering surface to be friction less
thus will do work on sled
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.
