Question

a student uses a force sensor to apply a constant net force to a 2.0 kg cart. another student measures its acceleration with a motion sensor. a model of the experiment is shown below. image of experiment setup the students repeat their experiment multiple times. they use the same cart but apply different strength net forces. they graph their data, which models how the cart’s acceleration changes as the net force changes. chart of acceleration vs. net force analyze the graph to complete the statements. - when the net force doubles, the acceleration dropdown - if the force is increased to 6 n, the acceleration will be blank m/s². show calculator

Explanation:

Step1: Determine the slope (acceleration per unit force)

The slope of the \(a\)-\(F\) graph is \(\frac{\Delta a}{\Delta F}\). Using a point like \((F = 4.0\,\text{N}, a = 2.0\,\text{m/s}^2)\), slope \(m_{\text{slope}}=\frac{2.0\,\text{m/s}^2}{4.0\,\text{N}} = 0.5\,\text{m/(s}^2\text{·N)}\).

Step2: Calculate acceleration for \(F = 6\,\text{N}\)

Using \(a = (\text{slope}) \times F\), substitute \(F = 6\,\text{N}\) and slope \(0.5\,\text{m/(s}^2\text{·N)}\):
\(a = 0.5\,\text{m/(s}^2\text{·N)} \times 6\,\text{N} = 3.0\,\text{m/s}^2\).

From the graph, we observe the relationship between net force (\(F\)) and acceleration (\(a\)). The graph is a straight line through the origin, indicating a direct proportionality (\(a \propto F\)) as per Newton's second law (\(F = ma\), where \(m\) is mass, constant here). For example, when \(F = 1.0\,\text{N}\), \(a = 0.5\,\text{m/s}^2\); when \(F = 2.0\,\text{N}\) (double of \(1.0\,\text{N}\)), \(a = 1.0\,\text{m/s}^2\) (double of \(0.5\,\text{m/s}^2\)). So when net force doubles, acceleration doubles.

Step1: Determine the slope (acceleration per unit force)

The slope of the \(a\)-\(F\) graph is \(\frac{\Delta a}{\Delta F}\). Using a point like \((F = 4.0\,\text{N}, a = 2.0\,\text{m/s}^2)\), slope \(m_{\text{slope}}=\frac{2.0\,\text{m/s}^2}{4.0\,\text{N}} = 0.5\,\text{m/(s}^2\text{·N)}\).

Step2: Calculate acceleration for \(F = 6\,\text{N}\)

Using \(a = (\text{slope}) \times F\), substitute \(F = 6\,\text{N}\) and slope \(0.5\,\text{m/(s}^2\text{·N)}\):
\(a = 0.5\,\text{m/(s}^2\text{·N)} \times 6\,\text{N} = 3.0\,\text{m/s}^2\).

Answer:

doubles

Second Part: If the force is increased to \(6\,\text{N}\), the acceleration will be...