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. 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. analyze the graph to complete the statements. - when the net force doubles, the acceleration dropdown. - if the force is increased to 6 n, the acceleratio text box m/s². options: is cut in half, stays the same, doubles, triples.

Explanation:

Step1: Analyze the graph's proportionality

The graph of acceleration vs. net force is a straight line through the origin, so acceleration \( a \) is directly proportional to net force \( F \) (i.e., \( a \propto F \)).

Step2: Apply proportionality to doubling force

If \( F \) doubles (e.g., from \( 1\,\text{N} \) to \( 2\,\text{N} \), or \( 2\,\text{N} \) to \( 4\,\text{N} \)), since \( a \propto F \), \( a \) must also double. For example, at \( F = 1\,\text{N} \), \( a = 0.5\,\text{m/s}^2 \); at \( F = 2\,\text{N} \), \( a = 1.0\,\text{m/s}^2 \) (doubled).

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

Explanation:

Step1: Find the slope (acceleration/force ratio)

From the graph, slope \( m=\frac{\Delta a}{\Delta F} \). Using \( F = 4\,\text{N} \), \( a = 2.0\,\text{m/s}^2 \), slope \( m=\frac{2.0}{4.0}=0.5\,\text{m/s}^2\text{ per N} \).

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

Using \( a = m \cdot F \), substitute \( m = 0.5\) and \( F = 6\):
\( a = 0.5 \times 6 = 3.0\,\text{m/s}^2 \).
Check proportionality: At \( F = 2\,\text{N} \), \( a = 1.0\,\text{m/s}^2 \); at \( F = 6\,\text{N} \) (3×2), \( a = 3×1.0 = 3.0\,\text{m/s}^2 \) (tripled from \( 2\,\text{N} \), but the key is the ratio: \( \frac{6}{4}=\frac{3}{2} \)? Wait, no—wait, initial at \( F = 4\,\text{N} \), \( a = 2.0 \). So \( 6\,\text{N} \) is \( \frac{6}{4}=1.5 \) times \( 4\,\text{N} \)? No, better to use the slope. Since \( a = \frac{F}{m_{\text{cart}}} \) (Newton’s second law, \( F = ma \Rightarrow a = F/m \)). The cart’s mass \( m = 2.0\,\text{kg} \), so \( a = F/2 \). For \( F = 6\), \( a = 6/2 = 3.0\,\text{m/s}^2 \).

Answers:

- When the net force doubles, the acceleration \(\boldsymbol{\text{doubles}}\).
- If the force is increased to \( 6\,\text{N} \), the acceleration is \(\boldsymbol{3.0}\,\text{m/s}^2\) (triples from \( 2\,\text{N} \)’s \( 1.0 \), but calculation shows \( 3.0 \)).

Step1: Analyze the graph's proportionality

The graph of acceleration vs. net force is a straight line through the origin, so acceleration \( a \) is directly proportional to net force \( F \) (i.e., \( a \propto F \)).

Step2: Apply proportionality to doubling force

If \( F \) doubles (e.g., from \( 1\,\text{N} \) to \( 2\,\text{N} \), or \( 2\,\text{N} \) to \( 4\,\text{N} \)), since \( a \propto F \), \( a \) must also double. For example, at \( F = 1\,\text{N} \), \( a = 0.5\,\text{m/s}^2 \); at \( F = 2\,\text{N} \), \( a = 1.0\,\text{m/s}^2 \) (doubled).

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

Step1: Find the slope (acceleration/force ratio)

From the graph, slope \( m=\frac{\Delta a}{\Delta F} \). Using \( F = 4\,\text{N} \), \( a = 2.0\,\text{m/s}^2 \), slope \( m=\frac{2.0}{4.0}=0.5\,\text{m/s}^2\text{ per N} \).

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

Using \( a = m \cdot F \), substitute \( m = 0.5\) and \( F = 6\):
\( a = 0.5 \times 6 = 3.0\,\text{m/s}^2 \).
Check proportionality: At \( F = 2\,\text{N} \), \( a = 1.0\,\text{m/s}^2 \); at \( F = 6\,\text{N} \) (3×2), \( a = 3×1.0 = 3.0\,\text{m/s}^2 \) (tripled from \( 2\,\text{N} \), but the key is the ratio: \( \frac{6}{4}=\frac{3}{2} \)? Wait, no—wait, initial at \( F = 4\,\text{N} \), \( a = 2.0 \). So \( 6\,\text{N} \) is \( \frac{6}{4}=1.5 \) times \( 4\,\text{N} \)? No, better to use the slope. Since \( a = \frac{F}{m_{\text{cart}}} \) (Newton’s second law, \( F = ma \Rightarrow a = F/m \)). The cart’s mass \( m = 2.0\,\text{kg} \), so \( a = F/2 \). For \( F = 6\), \( a = 6/2 = 3.0\,\text{m/s}^2 \).

Answer:

s:

• When the net force doubles, the acceleration \(\boldsymbol{\text{doubles}}\).
• If the force is increased to \( 6\,\text{N} \), the acceleration is \(\boldsymbol{3.0}\,\text{m/s}^2\) (triples from \( 2\,\text{N} \)’s \( 1.0 \), but calculation shows \( 3.0 \)).