In a simple postal scale, used to weigh letters, a flat pan is supported by an o
ID: 1326756 • Letter: I
Question
In a simple postal scale, used to weigh letters, a flat pan is supported by an ordinary spring (inside the device), as shown in the figure. When you place a letter on the pan, it causes the pan to move downward against the spring, while a pointer reads the weight of the letter. You observe that with nothing on it, the pan is at equilibrium 45.0 mm above the body of the scale. When you place a 0.700 oz letter on the scale, the pan moves down to 42.9 mm above the body of the scale. Recall that 1 oz is equal to 28.34952 g. Part A Find the force constant of this spring, in N/cm. K = N/cm SubmitMy AnswersGive Up Part B By how much from its equilibrium position would a 2.70 oz letter depress the pan? h = cm
Explanation / Answer
from the Hook's law
F = kx
(a)
(0.700oz)(1lb/16oz)(1N/0.2248lb)=k(45 mm- 42.9 mm)
0.1946 N = k ( 2.1 mm)
k = 0.1946 N/ 0.21 cm
=0.926 N/cm
b)
from the Hook's law
F=kh
(2.7oz)(1lb/16oz)(1N/0.2248lb)=0.926 N/cm (h )
h =0.750N/(0.926 N/cm)
=0.81 cm