1. For the larger spring, use the spring constant to calculate how far each of the following masses would stretch the spring, if they don’t exceed it elastic limit: (1) 950g (2) 1250g (3) 1600g
(1) x = F/K x = 950g/39.1 g/cm x =24.3 cm
(2) x = F/K x =1250g/39.1 g/cm x =31.9 cm
(3) x = F/K x = 1600g/39.1 g/cm x = 40.9 cm
2. For each spring, calculate the percent discrepancy between your result for the average spring constant, and the slope of the graph that you plotted for that spring.
3. If 500 newtons stretches a Hooke’s Law spring .4 meters, what is its spring constant in newtons per meter? What force would be needed to stretch this spring a distance of 1.5 meters?
K = F/x
K = 500n/.4m
K = 1250 n/m
F = Kx
F = (1250n/m)(1.5m)
F = 1875 n