Question

newton’s second law inquiry lab
conclusion: how do mass and force affect the acceleration of an object?
claim:

• increasing the mass will type answer here the object’s acceleration.
• increasing the force will type answer here the object’s acceleration.

evidence:
part 1:
comparing different masses
average distance traveled by golf ball (cm) type answer here
average distance traveled by ping pong ball (cm) type answer here
part 2: comparing different forces
distance ruler is bent (cm) average distance (cm)
2 type answer here
4 type answer here
6 type answer here
reasoning:

• in comparing two objects with different masses, when the same amount of force is applied, the acceleration of the lighter object is much type answer here because it has less type answer here
• increasing the amount of force resulted in type answer here acceleration, when the mass was kept constant.

Explanation:

Newton's second law of motion states that \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. Rearranged for acceleration, \( a=\frac{F}{m} \). When mass (\( m \)) increases with force (\( F \)) constant, acceleration (\( a \)) decreases (since \( a \) and \( m \) are inversely proportional). When force (\( F \)) increases with mass (\( m \)) constant, acceleration (\( a \)) increases (since \( a \) and \( F \)) are directly proportional. For the reasoning part, using \( a=\frac{F}{m} \), with \( F \) constant, a lighter mass (\( m \)) gives a larger \( a \). And when \( m \) is constant, more \( F \) means more \( a \).

Answer:

• Claim:
• Increasing the mass will \(\text{decrease}\) the object’s acceleration.
• Increasing the force will \(\text{increase}\) the object’s acceleration.
• Reasoning:
• In comparing two objects with different masses, when the same amount of force is applied, the acceleration of the lighter object is much \(\text{larger}\) because it has less \(\text{mass}\).
• Increasing the amount of force resulted in \(\text{increased}\) acceleration, when the mass was kept constant.