Use Hooke's Law and the Physics P.I.R.A.T.E. to help you solve the following problems.
Physics P.I.R.A.T.E.
As you continue working on problem solving in physics, don't forget your problem solving techniques. You may not think these steps are necessary now, but as the problems get more complex, this organization is essential.
| P | Draw a picture. |
| I | Identify the given information. |
| R | Write an equation using variables (letters). |
| A | Do the algebra and show your work. |
| T | Think about your answer. Does it make sense? |
| E | Explain your answer. |
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| A 4.92 N mass hanging vertically from a spring stretches the spring 0.750 m from its equilibrium position. What is the magnitude of the spring constant of the spring? |
|
\(\mathsf{ F_{elastic} = 4.92 \text{ N}}\) \(\mathsf{ k = ? \text{ N/m} }\) \(\mathsf{ x = -0.750 \text{ m}}\) |
\(\mathsf{ F_{elastic} = -kx }\) | \(\mathsf{ F_{elastic} = -kx }\) \(\mathsf{ 4.92 \text{ N} = -k(-0.750 \text{ m}) }\) \(\mathsf{ k = -\frac{4.92 \text{ N}}{-0.750 \text{ m}} }\) \(\mathsf{ k = 6.56 \text{ N/m} }\) |
| A 2.49 N mass is hanging vertically from a spring with k=35 N/m. What is the distance that the spring stretches from its equilibrium position? |
|
\(\mathsf{ F_{elastic} = 2.49 \text{ N}}\) \(\mathsf{ k = 35 \text{ N/m} }\) \(\mathsf{ x = ? \text{ m}}\) |
\(\mathsf{ F_{elastic} = -kx }\) | \(\mathsf{ F_{elastic} = -kx }\) \(\mathsf{ 2.49 \text{ N} = -(35 \text{ N/m})x }\) \(\mathsf{ x = -\frac{2.49 \text{ N}}{35 \text{ N/m}} }\) \(\mathsf{ x = -0.071 \text{ m} }\) |
| A 1.9 kg mass oscillates on a spring with a spring constant k = 43 N/m. What is the maximum displacement (amplitude) of the mass? |
|
\(\mathsf{ F_{elastic} = (1.9 \text{ kg})(9.81 \text{ m/s}^2) = 18.6 \text{ N}}\) \(\mathsf{ k = 43 \text{ N/m} }\) \(\mathsf{ x = ? \text{ m}}\) |
\(\mathsf{ F_{elastic} = -kx }\) | \(\mathsf{ F_{elastic} = -kx }\) \(\mathsf{ 18.6 \text{ N} = -(43 \text{ N/m})x }\) \(\mathsf{ x = -\frac{18.6 \text{ N}}{43 \text{ N/m}} }\) \(\mathsf{ x = -0.43 \text{ m} }\) |
