When a force of 36 Newtons is applied to springs S1 and S2, the displacement of the springs is 6 centimeters and 9 cm, respectively.
What is the difference between the spring constants of the two springs? helpp
k = F/x
so find k for each spring, and subtract
k1 = 36/6cm = .
k2 = 36/9cn =
To find the difference between the spring constants of the two springs, you need to first understand Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed as F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, you have the force applied (36 N) and the displacements of the two springs (6 cm and 9 cm). To determine the spring constants, you can rearrange Hooke's Law as k = F / x.
For spring S1, the force applied is 36 N and the displacement is 6 cm (or 0.06 m). Therefore, the spring constant of S1 can be calculated as k1 = 36 N / 0.06 m = 600 N/m.
For spring S2, the force applied is still 36 N, but the displacement is 9 cm (or 0.09 m). Using the same formula, the spring constant of S2 can be calculated as k2 = 36 N / 0.09 m = 400 N/m.
To find the difference between the two spring constants, you can subtract one from the other: difference = k1 - k2 = 600 N/m - 400 N/m = 200 N/m.
Therefore, the difference between the spring constants of S1 and S2 is 200 N/m.