A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
1 answer:
Answer:
Explanation:
The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:
Where:
There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:
The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since , so:
Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:
Where in this case:
Therefore:
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Refer to the diagram shown.
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
Answer:
2.5 m
Explanation:
Weight of billboard worker = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension in rope = 550 N
The sum of the torques will be
The position of the person will be 2.5 m