Answer:
0.615 m
Explanation:
We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.
The acceleration is determined using the equation of motion.
Given parameters:
Initial velocity, u = 0.00 m/s
Distance, s = 4.19 m
Time, t = 0.601 s
We use the equation
[tex]s = ut+\frac{1}{2}at^2[/tex]
With u = 0.00 m/s,
[tex]s = \frac{1}{2}at^2[/tex]
[tex]a = \dfrac{2s}{t^2}[/tex]
[tex]a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2[/tex]
The force is
[tex]F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}[/tex]
From Hooke's law, the extension, e, of a string is given by
[tex]e = \dfrac{F}{k}[/tex]
where k is the spring constant.
Hence,
[tex]e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}[/tex]
