contestada

A light spring with spring constant 1300 N/m hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant 1700 N/m. An object of mass 1.50 kg is hung at rest from the lower end of the second spring.(a) Find the total extension distance of the pair of springs.= m(b) Find the effective spring constant of the pair of springs as a system. We describe these springs as in series.= N/m

Respuesta :

Answer:

Part a)

[tex]x = 0.02 m[/tex]

Part b)

[tex]k = 737 N/m[/tex]

Explanation:

Let say the extension of spring 1 is x1 and extension of spring 2 is x2

then we will have

[tex]k_1x_1 = k_2x_2 = mg[/tex]

so we will have

[tex]x_1 = \frac{mg}{k_1}[/tex]

[tex]x_1 = \frac{1.50 \times 9.81}{1300}[/tex]

[tex]x_1 = 0.0113 m[/tex]

for other spring we will have

[tex]x_2 = \frac{mg}{k_2}[/tex]

[tex]x_2 = \frac{1.50 \times 9.81}{1700}[/tex]

[tex]x_2 = 0.00866 m[/tex]

so total extension of both the springs is given as

[tex]x = x_1 + x_2[/tex]

[tex]x = 0.0113 + 0.00866[/tex]

[tex]x = 0.02 m[/tex]

Part b)

Let say the effective spring constant is k

so we will have

[tex]kx = mg[/tex]

[tex]k = \frac{mg}{x}[/tex]

[tex]k = \frac{1.5 \times 9.81}{0.02}[/tex]

[tex]k = 737 N/m[/tex]

Otras preguntas