Question

courtney is playing with a spring toy. she holds the spring toy in her hand and lets one end of the spring toy drop down. gravity pulls the end of the spring toy downward with a force of 4 newtons. tension pulls the end of the spring toy upward with a force of 6 newtons. describe the spring toys motion. the end of the spring toy is while experiencing a net force. so, the magnitude of the spring toys velocity is .

Explanation:

Step1: Calculate Net Force

To find the net force, we subtract the downward force (gravity) from the upward force (tension) since they act in opposite directions. The formula for net force \( F_{net} \) is \( F_{net} = F_{upward} - F_{downward} \). Given \( F_{upward} = 6 \, \text{N} \) (tension) and \( F_{downward} = 4 \, \text{N} \) (gravity), we substitute these values: \( F_{net} = 6 - 4 = 2 \, \text{N} \) upward.

Step2: Analyze Motion Based on Net Force

According to Newton's second law, a net force causes acceleration (change in velocity). Since the net force is upward and the initial motion (from gravity pulling down) was downward, the toy is slowing down while moving downward (because the net force opposes the direction of motion initially) or speeding up if the net force is in the direction of motion. Wait, no—wait: the toy was let drop, so initial velocity is downward. The net force is upward (2 N), so the acceleration is upward. So if velocity is downward and acceleration is upward, the object is decelerating (slowing down) while moving downward, or if the net force is in the direction of future motion, but here initial motion is downward. Wait, no—let's clarify: when net force and velocity are in opposite directions, the object slows down. If net force and velocity are in the same direction, it speeds up. Here, velocity is downward (from being dropped), net force is upward (2 N), so they are opposite. So the toy is moving downward but slowing down (decelerating) because the net force is upward, opposing the downward motion. Alternatively, if the net force is upward, and the toy was moving downward, the acceleration is upward, so the velocity's magnitude (speed) is decreasing while moving downward, or if the toy was at rest, it would move upward, but here it was let drop, so initial motion is downward. Wait, maybe the first blank is "slowing down" (or "accelerating upward" but the context is motion direction). Wait, the first dropdown: possible options? Let's think again. The net force is upward (2 N). So the acceleration is upward. So if the toy is moving downward (initial velocity downward), then the acceleration is opposite to velocity, so it's slowing down (decelerating) in the downward direction. If the toy were moving upward, acceleration upward would speed it up, but here it's moving downward (from being dropped) initially. So the motion: the end of the spring toy is "slowing down" (or "decelerating") while experiencing a net force of "2 newtons upward" (or magnitude 2 N, direction upward). Then the magnitude of velocity is "decreasing" (since acceleration opposes velocity direction).

Wait, let's re-express:

- Net force: \( F_{net} = 6 - 4 = 2 \, \text{N} \) upward.

- Motion: Since the net force is upward, and the toy was moving downward (initial motion from being dropped), the acceleration is upward. So the toy is moving downward but slowing down (because acceleration is opposite to velocity direction). So the first blank: "slowing down" (or "decelerating"), net force: "2 newtons upward" (or magnitude 2 N), velocity magnitude: "decreasing".

But let's check the dropdowns. The first dropdown: options like "speeding up", "slowing down", "moving at constant speed"? Wait, the problem's dropdowns: let's assume the first blank is about the motion (direction and change in speed). The net force is 2 N upward. So if the toy is moving downward (initial velocity downward), the net force upward will cause it to slow down (decelerate) while moving downward, or if it comes to rest and then moves upward, but initially, it's moving downward. So the first part: "The end of the spring toy is [moving downward and slowing down] (or similar) while experiencing a net force of [2 newtons upward] (magnitude 2 N, direction upward). So the magnitude of the spring toy's velocity is [decreasing] (since acceleration is opposite to velocity, so speed decreases).

So putting it together:

1. First blank: The toy is moving downward (from being dropped) but slowing down (because net force is upward, opposing downward motion) → so "slowing down" (or "decelerating").

2. Net force: magnitude 2 N (6 - 4 = 2), direction upward → so "2 newtons upward" (or "2 N upward", or magnitude 2 N).

3. Velocity magnitude: since acceleration is opposite to velocity direction (downward velocity, upward acceleration), the speed (magnitude of velocity) is "decreasing".

Answer:

The end of the spring toy is \(\boldsymbol{\text{slowing down}}\) (or "decelerating") while experiencing a net force of \(\boldsymbol{2}\) newtons (magnitude, direction upward). So, the magnitude of the spring toy's velocity is \(\boldsymbol{\text{decreasing}}\).

(Note: The exact wording depends on the dropdown options, but the key is net force 2 N upward, motion: slowing down (if moving downward) or speeding up (if moving upward, but initial motion is downward), so slowing down, net force 2 N, velocity magnitude decreasing.)