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)

Question

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

Answer 1

To solve this question, we first need to calculate the original acceleration of the object when the net force is 0.8 N.

We can use Newton's second law of motion, which states:

\[ F = m \cdot a \]

where:

• \( F \) is the net force,
• \( m \) is the mass,
• \( a \) is the acceleration.

1. Calculate the original acceleration when the net force is 0.8 N:

\[ a_{\text{original}} = \frac{F}{m} = \frac{0.8 , \text{N}}{1.5 , \text{kg}} \approx 0.53 , \text{m/s}^2 \]

2. Now calculate the new acceleration when the net force is decreased to 0.2 N:

\[ a_{\text{new}} = \frac{F}{m} = \frac{0.2 , \text{N}}{1.5 , \text{kg}} \approx 0.13 , \text{m/s}^2 \]

3. Determine how the acceleration changed:

To find how the acceleration changed, we compare the new acceleration \( a_{\text{new}} \approx 0.13 , \text{m/s}^2 \) to the original acceleration \( a_{\text{original}} \approx 0.53 , \text{m/s}^2 \).

The new acceleration \( a_{\text{new}} \approx 0.13 , \text{m/s}^2 \) is about a quarter of the original acceleration which was \( 0.53 , \text{m/s}^2 \).

Thus, the correct response is:

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