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 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 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 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 under different net forces, we can use Newton's second law of motion, which states:

\[ F_{\text{net}} = m \cdot a \]

where:

• \( F_{\text{net}} \) is the net force,
• \( m \) is the mass of the object,
• \( a \) is the acceleration.

1. Calculating original acceleration with a net force of 0.8 N: \[ a = \frac{F_{\text{net}}}{m} = \frac{0.8 , \text{N}}{1.5 , \text{kg}} \approx 0.533 , \text{m/s}^2 \]

2. Calculating new acceleration with a decreased net force of 0.2 N: \[ a' = \frac{F_{\text{net}}'}{m} = \frac{0.2 , \text{N}}{1.5 , \text{kg}} \approx 0.133 , \text{m/s}^2 \]

Now, we can compare the new acceleration to the original acceleration:

• The original acceleration was approximately \( 0.533 , \text{m/s}^2 \).
• The new acceleration is \( 0.133 , \text{m/s}^2 \).

To quantify the change: \[ \text{Ratio of new acceleration to original acceleration} = \frac{0.133}{0.533} \approx \frac{1}{4} \]

Thus, the new acceleration is a quarter of the original value. Therefore, the correct response is:

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