Respuesta :
Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
The total energy of the system is 0.009 J
The maximum speed of the object is 0.19 m/s
The kinetic and potential energies of the system when the displacement is 2.00 cm are 0.005 J and 0.004 J respectively
The object is under simple harmonic motion.
Given that the mass of the object m =0.5 kg , spring constant k = 20N/m
(a) The total energy of the system in SMH,
[tex]E=\frac{1}{2}kA^2[/tex]
If the amplitude of motion A = 3cm = 0.03m, then total energy
[tex]E=\frac{1}{2}*20*0.03*0.03 J\\\\E = 0.009J[/tex]
E = 0.009J is the total energy of the system.
(b) maximum speed of the object
the maximum kinetic energy of the object = total energy of the object
[tex]\frac{1}{2} m(v_{max})^{2} =E\\\\v_{max}=\sqrt{2E/m}\\\\v_{max}=\sqrt{2*0.009/0.5}\\\\v_{max}=0.19m/s[/tex]
0.19 m/s is the maximum speed
(c) KE and PE when x = 2cm = 0.02m
potential energy at x = 0.02
[tex]PE=\frac{1}{2}kx^2\\\\PE=\frac{1}{2}*20*0.02*0.02\\\\PE= 0.004J[/tex]
Now, Kinetic energy + potential energy = total energy
KE + PE = E
KE = E - PE
KE = 0.009J - ).004J
KE = 0.005J
The kinetic and potential energies are 0.005 J and 0.004 J respectively.
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