That AY-Houses Hooker’s Law Lab Date conducted: September 18th 2013 Partners: Mohammed, Lima, Ben Unguent Purpose: The purpose of this experiment is to find the relationship between the stretch of the spring and the attached mass and to verify that this relationship is accurately described by Hook’s Law. Theory: Hooker’s Law states that to extend a spring by an amount (Stretch) from its previous position, one needs to add an external force (Mass). Therefore in order to verify
Hooker’s Law, you must verify that the mass (M) and the stretch(X) of the spring are proportional to each other. Apparatus: I used a: 1. Pole stand 2. A clamp 3. A spring 4. 5 (0. 050 keg) (Mass) 5. Metric Stick 6. Paper (For recording data) 7. Pen 8. Graph paper Procedure: Step 1: Collected all the materials needed to function this experiment Step 2: hang a Step 3: Calculated the length of the spring without any external force added to it. Step 4: Add 50 g to the spring and record the new length of the spring.
Need essay sample on Hooks law lab ?We will write a custom essay samplespecifically for you for only $13.90/pageorder now
Step 5: Add another egg to the spring and recorded the new length. Step 6: Repeat step 5 until the total mass added to the spring reached egg. Data: m/keg (Mass) 0. 050 0. 100 0. 150 0. 200 0. 250 mm (length) 0. 05 0. 07 0. 085 0. 105 0. 12 0. 14 Whom (Stretch) 0. 02 0. 035 0. 055 0. 09 Ex. Of how to find Stretch = 0. Mm = 0. 035 m Analysis: Finding the Equation: 0. Mm – 0. Mm 0. 15 keg- 0. 1 keg 0. 05 keg Discussion: – 0. Mm 0. 02 m Slope: 0. OMG-1 xx – XSL Equation: X= (0. OMG-1) (m)
Yes my experiment agreed with the theory Possibilities of error include: The ruler sliding from its position thus altering the length being measured. The ruler was old and round on the edges so it was hard to balance it on the table If the mass being added was not the mass it’s supposed to be. Ways to make improve this experiment: Better rulers A base that comes with a metric stick Conclusion: I have concluded that: X= (0. 4 meg-l) (m) And that: x o: m = Linear Therefore the stretch (x) is proportional to the mass (m)