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Sagot :
Given data:
* The mass of the marble is m = 0.8 kg.
* The spring constant is k = 625 N/m.
* The uncompressed length of the spring is L= 0.75 m.
* The compressed length of spring is L' = 0.4 m.
* The net mechanical energy of the system is E = 85 J.
* The value of time given is t = 6 s.
Solution:
(a). The change in the length of spring during the compression is,
[tex]\begin{gathered} x=L-L^{\prime} \\ x=0.75-0.4 \\ x=0.35\text{ m} \end{gathered}[/tex]The potential energy of the marble as it leaves the gun is,
[tex]U=\frac{1}{2}kx^2[/tex]Substituting the known values,
[tex]\begin{gathered} U=\frac{1}{2}\times625\times0.35^2 \\ U=38.28\text{ J} \end{gathered}[/tex]The kinetic energy of the marble as it leaves the gun in terms of the total mechanical energy is,
[tex]\begin{gathered} K=E-U \\ K=85-38.28 \\ K=46.72\text{ J} \end{gathered}[/tex]The kinetic energy of the marble in terms of mass and velocity of the marble is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ 46.72=\frac{1}{2}\times0.8\times v^2 \\ 46.72=0.4\times v^2 \\ v^2=\frac{46.72}{0.4} \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v^2=116.8 \\ v=10.8\text{ m/s} \end{gathered}[/tex]Thus, the velocity of the marble as it leaves the gun is 10.8 meters per second or approximately 11 meters per second.
(b). The force acting on the marble at the time of leave is,
[tex]\begin{gathered} F=kx \\ F=625\times0.35 \\ F=218.75\text{ N} \end{gathered}[/tex]According to Newton's second law, the force acting on the marble in terms of the acceleration of the marble is,
[tex]\begin{gathered} F=ma \\ 218.75=0.8\times a \\ a=\frac{218.75}{0.8} \\ a=273.44ms^{-2} \end{gathered}[/tex]By the kinematics equation, the distance traveled by the marble in 6 seconds is,
[tex]S=ut+\frac{1}{2}at^2[/tex]where u is the initial velocity of the marble (when the marble is at rest in the compressed spring position),
The value of the initial velocity of the marble is u = 0 m/s.
Substituting the known values,
[tex]\begin{gathered} S=0+\frac{1}{2}\times273.44\times6^2 \\ S=4921.92\text{ m} \end{gathered}[/tex]Thus, the distance traveled by the marble in 6 s is 4921.92 meters or approximately 4922 meters.
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