Question

intentional practice
table 2: acceleration vs mass

based on table 2, is the relationship between mass and acceleration directly or inversely proportional? explain with evidence from the data table.

Explanation:

When the net force is constant (here \(F = 50N\)), as the mass \(m\) increases:

• For \(m_1=5.0kg\), \(a_1 = 10m/s^{2}\), and \(m_1\times a_1=5.0\times10 = 50\)
• For \(m_2 = 10kg\), \(a_2=5.0m/s^{2}\), and \(m_2\times a_2=10\times5.0 = 50\)
• For \(m_3=15kg\), \(a_3 = 3.3m/s^{2}\), and \(m_3\times a_3\approx15\times3.3 = 49.5\approx50\)
• For \(m_4=20kg\), \(a_4=2.5m/s^{2}\), and \(m_4\times a_4=20\times2.5=50\)
• For \(m_5 = 25kg\), \(a_5=2.0m/s^{2}\), and \(m_5\times a_5=25\times2.0 = 50\)

According to Newton's second law \(F = ma\) (where \(F\) is constant), when one variable (mass \(m\)) increases, the other variable (acceleration \(a\)) decreases in such a way that their product (\(ma\)) remains constant. This is the characteristic of an inverse - proportional relationship (\(y=\frac{k}{x}\), here \(a=\frac{F}{m}\) with \(F = k = 50N\) constant)

Answer:

The relationship between mass and acceleration is inversely proportional. As shown in the data table, when the net force (\(F = 50N\)) is constant, \(m\times a\approx50\) (e.g., \(5.0\times10=50\), \(10\times5.0 = 50\), \(15\times3.3\approx50\), \(20\times2.5 = 50\), \(25\times2.0=50\)). According to \(F=ma\) (Newton's second law), when \(F\) is constant, \(a=\frac{F}{m}\), which is an inverse - proportional relationship.