Question

A spring stretches 5.52 cm vertically when a 2.50 kg object is suspended from it. Find the distance (in cm) the spring stretches if you replace the 2.50 kg object with a 3.40 kg object. Assume that the spring obeys the Hook's law.

Answer 1

Answer:

7.51 cm

Explanation:

y1 = 5.52 cm

m1 = 2.5 kg

m2 = 3.4 kg

Let the spring stretches by the distance y2.

According to Hooke's law

F = - k y

So, $\frac{F_{1}}{F_{2}}=\frac{y_{1}}{y_{2}}$

$\frac{2.5 \times 9.8}{3.4 \times 9.8}}=\frac{5.52}{y_{2}}$

y2 = 7.51 cm

Thus, the spring is stretches by 7.51 cm.