(a) 3539 N/m
Hook's law states that:
[tex]F=kx[/tex]
where F is the force applied on the spring, k is the spring constant, x is the stretching of the spring.
In this problem, we have:
[tex]F=mg=(65.0 kg)(9.8 m/s^2)=637 N[/tex] is the force applied (the weight of the fish)
[tex]x=0.180 m[/tex] is the stretching of the spring
Solving the equation for k, we find the spring constant:
[tex]k=\frac{F}{x}=\frac{637 N}{0.180 m}=3539 N/m[/tex]
(b) 0.85 s
The period of oscillation of a spring-mass system is
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
In this case,
m = 65.0 kg
k = 3539 N/m
Substituting into the formula,
[tex]T=2 \pi \sqrt{\frac{65.0 kg}{3539 N/m}}=0.85 s[/tex]
(c) 0.37 m/s
The initial elastic potential energy of the spring when the fish is pulled down is:
[tex]U=\frac{1}{2}k\Delta x^2[/tex]
where
[tex]\Delta x = 5.00 cm=0.05 m[/tex] is the stretching of the spring with respect to the initial position
Substituting,
[tex]U=\frac{1}{2}(3539 N/m)(0.05 m)^2=4.4 J[/tex]
The spring reaches its maximum speed when it crosses the equilibrium position, for which [tex]\Delta x=0[/tex], so when all the elastic potential energy has been converted into kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where v is the speed of the fish. Solving for v, we find
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(4.4 J)}{65.0 kg}}=0.37 m/s[/tex]
The force constant of the spring is 3.53 * 10^3 N/m, the period of oscillation of the fish is 0.115 s and the maximum speed it will reach is 0.869 m/s
A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m.
By using the amount of the spring is stretched by the weight of the fish, we can calculate the force constant [tex]k[/tex] of the spring.
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex] with [tex]v_{max}=\omega A = 2 \pi f A[/tex]
When the fish hangs at rest the upward spring force [tex]|F_x| = kx[/tex] equals the weight [tex]mg[/tex]of the fish [tex]f = \frac{1}{T}[/tex]. Therefore the amplitude of the SHM is [tex]0.0500 m.[/tex]
a)Find the force constant of the spring. Express your answer with the appropriate units.
[tex]mg=kx\\k = \frac{mg}{x} \\k = \frac{65*9.8}{0.180} = 3.53 * 10^3 N/m\\[/tex]
b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units.
[tex]T = 2\pi \sqrt{\frac{m}{k}[/tex]
[tex]T = 2\pi \sqrt{\frac{65 kg}{ 3.53*10^3 N/m} [/tex]
[tex]T =0.115 s[/tex]
c)What is the maximum speed it will reach?
[tex]v_{max}=\omega A = 2 \pi f A\\v_{max}=\omega A = \frac{ 2 \pi A}{T} \\v_{max}=\omega A = \frac{ 2 \pi 0.05 m}{0.115}\\v_{max}=\omega A = 0.869 m/s[/tex]
Learn more about
#LearnWithBrainly
