Asked by sdf
Newton's Second Law of Motion
Nash is investigating the relationship between the net force on an object and the acceleration of the object. He uses a spring scale to measure the force that is used to pull a cart along a track. He is able to change the mass of the cart by stacking steel plates on top of the cart. Nash sets up two photogates along the track. The photogates are used to accurately record the exact time the cart passes in front of them. This information can be used to calculate the average speed of the cart at and between the two photogates. The setup of the experiment is shown below.

Constructed Response Question
Nash collected the data shown in the table using the experimental setup.

A. Calculate the acceleration of the cart during Test 3 as described in the table. Show all your work and provide units in your answer.
B. Make a claim about the relationship between the net force on an object and the acceleration of the object that is supported by data in the table.
All Answers 1
Answered by
GPT-5 mini
AI
I’m missing the numeric values from the table/images you mentioned, so I can’t compute Test 3’s acceleration with the actual numbers. Please paste the Test 3 values (times, distances, or the speeds at the two photogates) and I’ll compute it exactly.
Meanwhile, here is the clear method you should use (and an example calculation) so you can follow or check the work.
A. How to calculate the acceleration for Test 3
- If the table gives the speed at the first photogate (v1), the speed at the second photogate (v2), and the time between the two photogates Δt (or the times the cart broke each gate so Δt = t2 − t1), then average acceleration a over that interval is
a = (v2 − v1) / Δt.
- Units: speeds in meters per second (m/s), time in seconds (s), so acceleration is m/s^2.
Example (illustration only)
- Suppose Test 3 gives v1 = 0.50 m/s, v2 = 0.80 m/s, and Δt = 0.60 s.
a = (0.80 − 0.50) / 0.60 = 0.30 / 0.60 = 0.50 m/s^2.
- So the acceleration for that example would be 0.50 m/s^2.
If instead the table gives only distances between photogates (Δx) and the times to pass each gate (t1, t2), you can:
- compute the average speed between gates vavg = Δx / (t2 − t1),
- or if the table gives times when the cart triggered each gate and you know the instantaneous speeds at the gates (or can infer them), use the same a = (v2 − v1) / Δt formula.
Provide the Test 3 numbers and I will show the full calculation with units.
B. Claim about relationship between net force and acceleration (supported by Newton’s Second Law)
- Claim: For a given (constant) mass, the acceleration of an object is directly proportional to the net force acting on it (i.e., if you double the net force, the acceleration doubles). Equivalently, the net force equals mass times acceleration: Fnet = m a.
- How to support this with your table data: hold mass constant and show that Fnet and a increase together proportionally (you can compute Fnet / a for each trial; it should be approximately the same value, equal to the mass). Or show that doubling Fnet roughly doubles a in trials with the same mass.
If you paste the table values I will compute Test 3 numerically and write a short claim that cites the numbers from your table.
Meanwhile, here is the clear method you should use (and an example calculation) so you can follow or check the work.
A. How to calculate the acceleration for Test 3
- If the table gives the speed at the first photogate (v1), the speed at the second photogate (v2), and the time between the two photogates Δt (or the times the cart broke each gate so Δt = t2 − t1), then average acceleration a over that interval is
a = (v2 − v1) / Δt.
- Units: speeds in meters per second (m/s), time in seconds (s), so acceleration is m/s^2.
Example (illustration only)
- Suppose Test 3 gives v1 = 0.50 m/s, v2 = 0.80 m/s, and Δt = 0.60 s.
a = (0.80 − 0.50) / 0.60 = 0.30 / 0.60 = 0.50 m/s^2.
- So the acceleration for that example would be 0.50 m/s^2.
If instead the table gives only distances between photogates (Δx) and the times to pass each gate (t1, t2), you can:
- compute the average speed between gates vavg = Δx / (t2 − t1),
- or if the table gives times when the cart triggered each gate and you know the instantaneous speeds at the gates (or can infer them), use the same a = (v2 − v1) / Δt formula.
Provide the Test 3 numbers and I will show the full calculation with units.
B. Claim about relationship between net force and acceleration (supported by Newton’s Second Law)
- Claim: For a given (constant) mass, the acceleration of an object is directly proportional to the net force acting on it (i.e., if you double the net force, the acceleration doubles). Equivalently, the net force equals mass times acceleration: Fnet = m a.
- How to support this with your table data: hold mass constant and show that Fnet and a increase together proportionally (you can compute Fnet / a for each trial; it should be approximately the same value, equal to the mass). Or show that doubling Fnet roughly doubles a in trials with the same mass.
If you paste the table values I will compute Test 3 numerically and write a short claim that cites the numbers from your table.
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