Respuesta :
Answer:
(a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
Explanation:
Given that,
Spring stiffness = 205 N/m
Mass = 0.6 kg
Compression of spring = 13 cm
Initial speed = 3 m/s
(a). We need to calculate the maximum stretch during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2[/tex]
[tex]x_{m}=\sqrt{\dfrac{4.43\times2}{205}}[/tex]
[tex]x_{m}=20.7\ cm[/tex]
(b). Maximum speed comes when stretch is zero.
We need to calculate the maximum speed during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2[/tex]
[tex]v'=\sqrt{\dfrac{4.43\times2}{0.6}}[/tex]
[tex]v'=3.84\ m/s[/tex]
(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.6}{205}}[/tex]
[tex]T=0.33\ sec[/tex]
We need to calculate the energy
Using formula of energy
[tex]E=\dfrac{P}{t}[/tex]
Put the value into the formula
[tex]E=\dfrac{0.02}{0.33}[/tex]
[tex]E=0.060\ Watt[/tex]
Hence, (a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
