contestada

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?

Respuesta :

Answer:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

[tex]mgx - \frac{1}{2}kx^2 = 0[/tex]

[tex](2.07)(9.8)(0.131) - \frac{1}{2}k(0.131)^2 = 0[/tex]

now we will have

[tex] k = 310 N/m[/tex]

Part B)

Time period of oscillation is given as

[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi\sqrt{\frac{2.07}{310}}[/tex]

[tex]T = 0.51 s[/tex]

Otras preguntas