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)
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 0.13 m/s2.
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 0.13 m/s2.
original a = F/m = .8/1.5 = .5333...
final a = .53333... / 4 = .13333.......
So which answer is that?
To determine what happens to the acceleration when the net force is decreased, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force and inversely proportional to its mass. The formula for calculating acceleration is:
a = Fnet / m
where a is the acceleration, Fnet is the net force, and m is the mass of the object.
In the given scenario, the net force is initially 0.8 N, and the mass is 1.5 kg. So, using the formula, we can find the initial acceleration:
a_initial = 0.8 N / 1.5 kg
Simplifying the calculation gives:
a_initial = 0.533 m/s^2
Now, let's see what happens when the net force is reduced to 0.2 N. We can use the same formula to calculate the new acceleration:
a_new = 0.2 N / 1.5 kg
Simplifying the calculation gives:
a_new = 0.133 m/s^2
Comparing the initial and new accelerations, we can see that the new acceleration is approximately one-fourth (or a quarter) of the initial acceleration. Therefore, the correct answer is:
The acceleration decreases to a quarter of its original value, or about 0.13 m/s^2.