Answer:
Explanation:
There seems to be a typo in the problem statement. It says the spring stretches to a shorter length after more mass is added. Please check the problem statement. I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.
Hooke's law states that the force needed to compress or extend a linear spring is:
F = kΔx, where k is the stiffness and Δx is the displacement.
When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm. The force of the spring is counteracting the weight of both the pan and the object. Therefore:
(m + 0.020) g = k (0.1180 - 0.100)
And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.
(m + 0.020 + 0.030) g = k (0.1205 - 0.100)
We now have two equations and two variables. If we divide the second equation by the first equation:
(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)
(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180
0.0180 (m + 0.050) = 0.02050 (m + 0.020)
0.0180 m + 0.0009 = 0.02050 m + 0.00041
0.00049 = 0.0025 m
m = 0.196
The pan has a mass of 0.196 kg, or 196 g.
