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? Remember F= M*A
a The acceleration increases to four times its original value, or about 2.13 m/s2.
b The acceleration increases to four times its original value, or about 0.13 m/s2.
c The acceleration decreases to a quarter of its original value, or about 2.13 m/s2.
d The acceleration decreases to a quarter of its original value, or about 0.13 m/s2.
To find the answer, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. The equation is given as F = m * a, where F is the net force, m is the mass, and a is the acceleration.
Given that the net force is initially 0.8 N and the mass is 1.5 kg, we can rearrange the equation to solve for acceleration: a = F / m. Plugging in the values, we have a = 0.8 N / 1.5 kg, which gives us an initial acceleration of approximately 0.53 m/s^2.
Now, let's consider what happens when the net force is decreased to 0.2 N. Using the same equation, we can find the new acceleration: a = 0.2 N / 1.5 kg. This gives us a new acceleration of approximately 0.13 m/s^2.
Comparing the initial acceleration of 0.53 m/s^2 with the new acceleration of 0.13 m/s^2, we can see that the acceleration decreases to a quarter (one-fourth) of its original value. Therefore, the correct answer is option d: The acceleration decreases to a quarter of its original value, or about 0.13 m/s^2.