Question

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

Answer 1

Answer:

Part a)

$x = 0.02 m$

Part b)

$k = 737 N/m$

Explanation:

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

then we will have

$k_1x_1 = k_2x_2 = mg$

so we will have

$x_1 = \frac{mg}{k_1}$

$x_1 = \frac{1.50 \times 9.81}{1300}$

$x_1 = 0.0113 m$

for other spring we will have

$x_2 = \frac{mg}{k_2}$

$x_2 = \frac{1.50 \times 9.81}{1700}$

$x_2 = 0.00866 m$

so total extension of both the springs is given as

$x = x_1 + x_2$

$x = 0.0113 + 0.00866$

$x = 0.02 m$

Part b)

Let say the effective spring constant is k

so we will have

$kx = mg$

$k = \frac{mg}{x}$

$k = \frac{1.5 \times 9.81}{0.02}$

$k = 737 N/m$