Question

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

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

Answer 1

To find the acceleration of the object in both scenarios, we can use Newton's second law of motion, which states that \( F_{net} = m \cdot a \), where \( F_{net} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

1. **Initial scenario**:
- Net force \( F_{net} = 0.8 \, \text{N} \)
- Mass \( m = 1.5 \, \text{kg} \)

We can rearrange the equation to solve for acceleration:
\[
a = \frac{F_{net}}{m} = \frac{0.8 \, \text{N}}{1.5 \, \text{kg}} \approx 0.5333 \, \text{m/s}^2.
\]

2. **New scenario**:
- Net force \( F_{net} = 0.2 \, \text{N} \)

Again, calculate the acceleration:
\[
a = \frac{F_{net}}{m} = \frac{0.2 \, \text{N}}{1.5 \, \text{kg}} \approx 0.1333 \, \text{m/s}^2.
\]

Now, let's compare the two accelerations:
- Initial acceleration \( a_1 \approx 0.5333 \, \text{m/s}^2 \)
- New acceleration \( a_2 \approx 0.1333 \, \text{m/s}^2 \)

Notice that the new acceleration \( \approx 0.1333 \, \text{m/s}^2 \) is indeed \( \frac{1}{4} \) of the original acceleration \( \approx 0.5333 \, \text{m/s}^2 \).

Therefore, the correct response is:
**The acceleration decreases to a quarter of its original value, or about 0.13 m/s².**