k{\displaystyle k}: Spring constant F{\displaystyle F}: Force (the force applied to the spring) x{\displaystyle x}: Distance the spring is compressed or stretched away from its equilibrium or rest position
The negative symbol indicates that the force of the spring constant is in the opposite direction of the force applied to the spring. It does not indicate that the value is negative. Hooke’s law is actually pretty limited. It only applies to perfectly elastic materials within their elastic limit—stretch something too far and it’ll break or stay stretched out. Hooke’s law is based on Newton’s third law of motion, which states that for every action there is an equal and opposite reaction. [4] X Research source
For example, if you cut a spring in half, its spring constant will double. If you doubled the length of the spring, on the other hand, its spring constant would be half what it was. This also means that when you apply the same force to a longer spring as a shorter spring, the longer spring will stretch further than the shorter spring. [6] X Research source
Hint: Since the spring constant is measured in N/m{\displaystyle N/m}, convert 3. 5cm{\displaystyle 3. 5cm} to 0. 035m{\displaystyle 0. 035m}.
Hint: Use the equation for Hooke’s Law without the negative symbol, which doesn’t have a mathematical function: F=kx{\displaystyle F=kx}.
Hint: The person’s weight is equivalent to the force applied to the spring.
Hint: Remember to convert the cm{\displaystyle cm} to m{\displaystyle m} since the spring constant is measured in N/m{\displaystyle N/m}.
The formula to find the spring constant is k=Fx{\displaystyle k={\frac {F}{x}}}. Here, the force is 0. 1N{\displaystyle 0. 1N} and the distance the spring stretches when that force is added is 0. 035m{\displaystyle 0. 035m}, so your equation is k=0. 10. 035{\displaystyle k={\frac {0. 1}{0. 035}}}. Divide 0. 1{\displaystyle 0. 1} by 0. 035{\displaystyle 0. 035}, then round to get your answer of 2. 85N/m{\displaystyle 2. 85N/m}. Why not 0. 2N{\displaystyle 0. 2N}? Because you don’t know the distance the string stretched when the first weight was applied. You only know the additional distance it stretched when the second weight was applied.
k=Fx{\displaystyle k={\frac {F}{x}}} k=6700. 0079{\displaystyle k={\frac {670}{0. 0079}}} k=84,810N/m{\displaystyle k=84,810N/m}
k=252. 8{\displaystyle k={\frac {25}{2. 8}}} k=8. 93N/m{\displaystyle k=8. 93N/m}
If you’re given a line that represents a spring that obeys Hooke’s Law (also called an ideal spring), you can find the spring constant by finding the slope of the line using the basic slope formula y2−y1x2−x1{\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}. A line with a spring constant as a slope will always cross through the origin of the graph.
