Question

A spring scale shows a net force of 0.8 N acting on a 1.5-kg mass. What happens to the acceleration of the object if the net force is decreased to 0.2 N?(1 point)
Responses

The acceleration decreases to a quarter of its original value, or about 2.13 m/s2.
The acceleration decreases to a quarter of its original value, or about 2.13 m/s 2 .

The acceleration decreases to a quarter of its original value, or about 0.13 m/s2.
The acceleration decreases to a quarter of its original value, or about 0.13 m/s 2 .

The acceleration increases to four times its original value, or about 0.13 m/s2.
The acceleration increases to four times its original value, or about 0.13 m/s 2 .

The acceleration increases to four times its original value, or about 2.13 m/s2.

Answer 1

To determine how the acceleration changes with different net forces acting on the object, we can use Newton's second law of motion, which states:

\[ F_{\text{net}} = ma \]

Where:
- \( F_{\text{net}} \) is the net force,
- \( m \) is the mass,
- \( a \) is the acceleration.

First, let's calculate the acceleration with the original net force of 0.8 N:

Given:
- \( F_{\text{net}} = 0.8 \, \text{N} \)
- \( m = 1.5 \, \text{kg} \)

Using the formula, we can solve for acceleration \( a \):

\[ a = \frac{F_{\text{net}}}{m} = \frac{0.8 \, \text{N}}{1.5 \, \text{kg}} = \frac{0.8}{1.5} \approx 0.533 \, \text{m/s}^2 \]

Now, let's calculate the acceleration with the new net force of 0.2 N:

Given:
- \( F_{\text{net}} = 0.2 \, \text{N} \)
- \( m = 1.5 \, \text{kg} \)

Again using the formula, we can solve for the new acceleration \( a \):

\[ a = \frac{F_{\text{net}}}{m} = \frac{0.2 \, \text{N}}{1.5 \, \text{kg}} = \frac{0.2}{1.5} \approx 0.133 \, \text{m/s}^2 \]

Now we can compare the two accelerations. The initial acceleration was approximately 0.533 m/s² and the new acceleration is approximately 0.133 m/s², which is indeed a quarter of the original acceleration:

\[
0.133 \, \text{m/s}^2 \approx \frac{0.533 \, \text{m/s}^2}{4}
\]

Therefore, the correct response is:

**The acceleration decreases to a quarter of its original value, or about 0.13 m/s².**