Answer:
0.5 m
14.00595
8 m/s, 0.0625 s
5.71314 m/s
Explanation:
k = Spring constant = 128 N/m
A = Amplitude
E = Energy in spring = 16 J
Energy in spring is given by
[tex]E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m[/tex]
The amplitude is 0.5 m
Time period is given by
[tex]T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s[/tex]
Number of oscillations is given by
[tex]N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595[/tex]
The number of oscillations is 14.00595
For maximum speed
[tex]\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s[/tex]
The maximum speed is 8 m/s
For a distance of 0.5 m which is the amplitude
[tex]Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s[/tex]
The time taken would be 0.0625 s
The maximum kinetic energy is equal to the mechanical energy
[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16[/tex]
At x = 0.35 m
[tex]v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s[/tex]
The speed of the block is 5.71314 m/s
Answer
given,
k = spring constant = 128 N/m
mass of block = 0.5 Kg
mechanical energy = 16 J
using formula
[tex]E = \dfrac{1}{2}kA^2[/tex]
[tex]16 = \dfrac{1}{2}\times 128 \times A^2[/tex]
A² = 0.25
A = 0.5 m
b) T = time period of one oscillation
[tex]T = 2\pi \sqrt{\dfrac{m}{k}}[/tex]
[tex]T = 2\pi \sqrt{\dfrac{0.5}{128}}[/tex]
T = 0.393 s
number of oscillations = [tex]\dfrac{5.5}{0.393}[/tex]
n = 14
c)
maximum K E = mechanical energy = 16 J
d) using conservation of energy
K E + E_{sp} = mechanical energy
[tex]\dfrac{1}{2}mv^2 +\dfrac{1}{2}kx^2 =16[/tex]
at x = 0.35 m
[tex]\dfrac{1}{2}\times 0.5\times v^2 +\dfrac{1}{2}\times 128 \times 0.35^2 =16[/tex]
v = 5.71 m/s
