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Do these results (see chart) verify Newton's second law? Explain why

Do these results (see chart) verify Newton's second law? Explain why

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Do these results (see chart) verify Newton's second law? Explain why 

**Table 4.2: Force and Acceleration** - **Mass on the Hanger (kg)** - 0.0108 - 0.0188 - 0.0148 - 0.0108 - **Net Force on the System (N)** - 0.1058 - 0.1842 - 0.1450 - 0.1058 - **Acceleration Trial 1 (m/s²)** - 0.2097 - 0.4107 - 0.4147 - 0.5132 - **Acceleration Trial 2 (m/s²)** - 0.3677 - 0.4148 - 0.5493 - 0.8153 - **Acceleration Trial 3 (m/s²)** - 0.3658 - 0.4054 - 0.5524 - 0.8135 - **Average Acceleration (m/s²)** - 0.3012 - 0.4103 - 0.5443 - 0.8237 **Total Mass of the System: 1.33 kg** **Note:** - The table outlines various measurements of force and acceleration. - A consistent notation is used for decimal representation, denoted by leading zeros. - A formula is hinted at the side: \( a = \frac{\text{F}_{\text{net}}}{\text{mass}} \), suggesting acceleration is calculated by dividing the net force by mass, and is considered a constant.
Transcribed Image Text:**Table 4.2: Force and Acceleration** - **Mass on the Hanger (kg)** - 0.0108 - 0.0188 - 0.0148 - 0.0108 - **Net Force on the System (N)** - 0.1058 - 0.1842 - 0.1450 - 0.1058 - **Acceleration Trial 1 (m/s²)** - 0.2097 - 0.4107 - 0.4147 - 0.5132 - **Acceleration Trial 2 (m/s²)** - 0.3677 - 0.4148 - 0.5493 - 0.8153 - **Acceleration Trial 3 (m/s²)** - 0.3658 - 0.4054 - 0.5524 - 0.8135 - **Average Acceleration (m/s²)** - 0.3012 - 0.4103 - 0.5443 - 0.8237 **Total Mass of the System: 1.33 kg** **Note:** - The table outlines various measurements of force and acceleration. - A consistent notation is used for decimal representation, denoted by leading zeros. - A formula is hinted at the side: \( a = \frac{\text{F}_{\text{net}}}{\text{mass}} \), suggesting acceleration is calculated by dividing the net force by mass, and is considered a constant.
Expert Solution
Step 1

Newton's Second Law

Newton's second law states that the acceleration of an object is directly proportional to the applied force and inversely proportional to the mass of the object

a=Fnetm

here mass of the object m = 0.33 kg

The experiment is performed on this object by varying the applied forces

 

 

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