HS-PS2-1: Analyzing Data to Support Newton's Second Law
In plain English: HS-PS2-1 is the NGSS high school performance expectation that asks students to analyze motion data to support the claim that Newton's second law, F = ma, links the net force on an object, its mass and its acceleration. Students find acceleration from position-time and velocity-time data and compare it with net force divided by mass. It is usually taught in Physics or Physical Science.
Analyze data to support the claim that Newton's second law of motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration.
Clarification Statement: Examples of data could include tables or graphs of position or velocity as a function of time for objects subject to a net unbalanced force, such as a falling object, an object rolling down a ramp, or a moving object being pulled by a constant force.
Assessment Boundary: Assessment is limited to one-dimensional motion and to macroscopic objects moving at non-relativistic speeds.
Next Generation Science Standards · Disciplinary Core Idea: Motion and Stability: Forces and Interactions · Official standard
Students use motion data to test the claim that the acceleration of an object equals the net force on it divided by its mass. They read acceleration as the slope of a velocity-time graph, recover it from position-time data for motion that starts from rest, and compare it with Fnet/m. The data come from carts pulled by a steady force, carts rolling down a ramp and falling objects.
The NGSS Clarification Statement says: "Examples of data could include tables or graphs of position or velocity as a function of time for objects subject to a net unbalanced force, such as a falling object, an object rolling down a ramp, or a moving object being pulled by a constant force." The Assessment Boundary says: "Assessment is limited to one-dimensional motion and to macroscopic objects moving at non-relativistic speeds." Every data set on this page is invented for teaching, uses g = 9.8 m/s², and describes motion along one line. Students do the practice of analyzing data: they compute, graph, compare with a prediction and write a claim backed by evidence.
Learning Objectives
By the end of this lesson, students will be able to:
Find the acceleration of an object as the slope of a velocity-time graph or table
Find the acceleration of an object that starts from rest from position-time data, using x = ½at²
Compare a measured acceleration with Fnet/m and judge whether the data support Newton's second law, allowing for measurement uncertainty
Use graphs of acceleration against net force (constant mass) and against mass (constant force) to describe the relationship among the three quantities
Write a claim, evidence and reasoning statement that uses data to support the claim that Fnet = ma
Prior Knowledge Required
Students should already be comfortable with:
Balanced and unbalanced forces and how they change motion MS-PS2-2
Speed, velocity and acceleration as rates of change, with units m/s and m/s²
Finding the slope of a line from a table or graph 8.F.B.4
Drawing a simple free-body diagram and adding forces along one line
Show two shopping carts, one empty and one loaded with groceries. Ask students to predict before any numbers appear:
Warm-Up Prompt
"You give the empty cart and the loaded cart the same steady push, starting from rest. Sketch both velocity-time graphs on one set of axes. What is the same about the two graphs, what is different, and what would you measure to test your sketch?"
Collect sketches. Many students draw two straight lines with different slopes; ask what the slope stands for. Record their ideas about what to measure (mass, force, time, distance) on the board: the lesson returns to them.
Direct Instruction15-20 minutes
Reading acceleration from data. Acceleration is the change in velocity per unit time, so on a velocity-time graph it is the slope. For an object that starts from rest with constant acceleration, position follows x = ½at², so a = 2x/t², and the second differences of equally spaced position data are equal. Newton's second law predicts a = Fnet/m, where Fnet is the sum of all forces along the line of motion. The test is to compare the acceleration measured from motion data with Fnet/m found from force and mass measurements. Use Diagram 1 to show two carts pulled by the same 0.80 N net force.
Cart pulled by a steady force (velocity-time)
A 0.50 kg cart on a low-friction track is pulled by a force sensor that reads 0.80 N. A motion detector records v = 0, 0.80, 1.60, 2.40, 3.20 m/s at t = 0, 0.5, 1.0, 1.5, 2.0 s.
Result: Slope = 3.20 m/s ÷ 2.0 s = 1.6 m/s². Prediction: Fnet/m = 0.80 N ÷ 0.50 kg = 1.6 m/s². The data agree.
Same force, different masses
A 1.20 N net force is applied to carts of mass 0.50, 1.00, 1.50 and 2.00 kg. The measured accelerations are 2.4, 1.2, 0.8 and 0.6 m/s².
Result: m × a = 1.20 N for every cart, so doubling the mass halves the acceleration: a is proportional to 1/m.
Position-time data from rest
A 2.0 kg cart starts from rest. Its position is x = 0, 0.15, 0.60, 1.35 m at t = 0, 1, 2, 3 s.
Result: First differences 0.15, 0.45, 0.75 m; second differences 0.30 m, so a = 0.30 m/s² (also 2x/t² = 2(1.35)/9 = 0.30). Fnet = ma = 0.60 N.
Falling object
A 0.20 kg steel ball is dropped. A motion detector records v = 0, 0.98, 1.96, 2.94, 3.92 m/s at t = 0, 0.1, 0.2, 0.3, 0.4 s (downward positive).
Result: Slope = 9.8 m/s². The only significant force is the weight mg = 1.96 N, and Fnet/m = 1.96 ÷ 0.20 = 9.8 m/s².
Rolling down a ramp
A 0.80 kg cart rolls from rest down a ramp: v = 0, 0.30, 0.60, 0.90 m/s at t = 0, 0.25, 0.50, 0.75 s. A spring scale parallel to the ramp needs 1.10 N to hold the cart still.
Result: a = 1.2 m/s², so Fnet = 0.80 × 1.2 = 0.96 N down the ramp. The 1.10 N pull of gravity along the ramp minus friction gives 0.96 N, so friction is about 0.14 N.
After each example, ask: "Which part of the data is the evidence, and what calculation turns it into a test of the claim?" Stress that the starting velocity does not affect the slope, and that forces opposite to the motion (friction, air resistance) make Fnet smaller than the applied force.
Guided Practice15 minutes
Pairs analyze the invented data below from one 0.75 kg cart (mass measured on a balance) pulled by four different net forces. They plot acceleration against net force, draw a best-fit line and interpret its slope. Then compare with Diagram 2.
Guided practice data: 0.75 kg cart
Trial
Net force F (N)
Measured a (m/s²)
1
0.30
0.41
2
0.60
0.79
3
0.90
1.21
4
1.20
1.59
Plot the four points with net force on the horizontal axis and acceleration on the vertical axis.
Draw the best-fit line. Does it pass close to the origin? What would a line that misses the origin by a lot suggest?
Find the slope. Newton's second law says a = (1/m)F, so the slope should be 1/m. Here the slope is about 1.33 per kg, so m ≈ 0.75 kg, matching the balance.
Write one sentence that states the claim and cites the slope as evidence.
Independent Practice10-15 minutes
Each student analyzes a new invented data set: a 1.5 kg cart is pulled along a level track while a force sensor reads a steady 1.8 N, and friction is negligible. The cart is already moving when recording starts.
Independent practice data: 1.5 kg cart, 1.8 N
t (s)
0
0.4
0.8
1.2
1.6
v (m/s)
0.10
0.58
1.06
1.54
2.02
Students find the acceleration (1.2 m/s²), compute Fnet/m (1.8 ÷ 1.5 = 1.2 m/s²), and write a claim, evidence and reasoning paragraph. Ask them to explain why the 0.10 m/s starting velocity does not change the result.
Closure5 minutes
Exit ticket: "A 0.40 kg cart's velocity rises steadily from 0.2 m/s to 1.4 m/s in 2.0 s. What net force acts on it? Name the evidence you used." (a = 0.6 m/s², Fnet = 0.24 N.) Return to the warm-up sketches and ask students to label each slope with Fnet/m.
Differentiation Strategies
For Struggling Students
Give a data table with an extra column for Δv and Δt so students compute the slope one interval at a time before finding the overall slope
Provide a three-row claim-evidence-reasoning frame: "The data show ___. The acceleration is ___ and Fnet/m is ___. These agree because ___."
Start with velocity-time data before position-time data, and use whole-number masses and forces for the first examples
For Advanced Students
Ask students to estimate the uncertainty in each measured acceleration and decide whether the difference from Fnet/m is larger than the uncertainty
Have students use the intercept of the acceleration-force graph to estimate the friction force on the cart
Ask students to design a data table that tests a ∝ 1/m while keeping the net force fixed, and explain which graph would come out as a straight line through the origin
Assessment Guidance
What to Look For
Check that students find acceleration from the slope, not from a single velocity divided by a single time when the object did not start from rest. Look for claims that name the data (a slope, a table value, the product m × a) and compare it with a prediction. Listen for students who treat the applied force as the net force when friction or air resistance acts, and for students who think a moving object must have a net force on it.
02
Classroom Activities
3 Activities
1
Constant Mass, Changing Force Lab
25 minGroups of 3-4
Groups pull a cart along a level track with a string that runs over a pulley to a hanging mass. They move slotted masses from the cart to the hanger, so the net force changes while the total moving mass stays at 1.00 kg. They time the cart over 1.00 m from rest and compute a = 2d/t².
Procedure
Set up the cart (0.92 kg), a very light hanger (a paper clip hook whose mass can be ignored) and eight 10 g slotted masses. Start with 2 on the hanger and 6 on the cart: total moving mass 1.00 kg.
Release the cart from rest, time it over 1.00 m (stopwatch or phone video at 30 or 60 frames per second), and repeat three times.
Move 2 more slotted masses to the hanger and repeat, up to 8 on the hanger. The pulling force is the weight of the hanging masses, mhg.
Safety: put a stop block at the end of the track, catch the cart by hand, and keep feet out from under the hanging mass.
Sample Data (invented)
Hanging mass 0.020, 0.040, 0.060, 0.080 kg (pulling force 0.196, 0.392, 0.588, 0.784 N). Average time over 1.00 m: 3.37, 2.32, 1.88 and 1.62 s. Students compute a = 2(1.00)/t² for each trial (about 0.176, 0.372, 0.566, 0.762 m/s²) and plot a against the pulling force.
Discussion Questions
The best-fit line has a slope close to 1.0 per kg. What does that tell you about the moving mass?
The line crosses the force axis at about 0.02 N instead of at zero. What force could explain this, and in which direction does it act?
Why do we move masses from the cart to the hanger instead of adding new ones?
Modification for Distance Learning
Students use a free cart-and-pulley simulation or analyze the sample data above in a spreadsheet, then post their graph and claim to the class discussion board.
2
Coffee Filter Drop: When the Net Force Is Zero
20 minPairs
Pairs drop a stack of four nested basket coffee filters (total mass 4.0 g) and record the fall with phone video next to a meter stick. They build a velocity-time table and use Newton's second law to reason about the forces at the start of the fall and after the speed stops changing.
Procedure
One partner holds the stack flat, open side up, about 2 m above the floor while standing on the floor or a sturdy step stool with a spotter (never a chair).
The other partner records video. Step through the frames and read the position every 0.1 s, then compute the velocity for each interval.
Repeat with a single filter and compare the final speeds.
The average acceleration in the first 0.1 s is 8.0 m/s², less than 9.8 m/s². What does that tell you about the net force at the start?
From 0.5 s to 0.7 s the velocity does not change. What is the net force then, and how large is the air resistance on the 4.0 g stack?
Why does a single filter reach a smaller final speed than the stack?
Challenge Variation
Groups predict, then test, whether a stack of nine filters reaches a larger final speed than the stack of four, and explain the prediction with forces.
3
Data Stations: Claim, Evidence, Reasoning
25 minGroups of 3-4
Four stations each hold one invented data set. At each station, groups find the acceleration, compute the net force with Fnet = ma, and write one claim-evidence-reasoning card. Groups rotate every 6 minutes.
Station Data
Station 1, hockey puck on ice (0.17 kg): v = 12.0, 11.7, 11.4, 11.1 m/s at t = 0, 0.5, 1.0, 1.5 s.
Station 2, car on a level test track (1,200 kg): v = 0, 3.0, 6.0, 9.0 m/s at t = 0, 2, 4, 6 s.
Station 3, tennis ball tossed straight up (0.058 kg, upward positive): v = 4.90, 2.94, 0.98, -0.98 m/s at t = 0, 0.2, 0.4, 0.6 s.
Station 4, fan cart from rest (1.2 kg, fan thrust measured at 0.30 N): x = 0, 0.125, 0.500, 1.125 m at t = 0, 1, 2, 3 s.
Discussion Questions
At which stations does the net force point opposite to the velocity for the whole data set? How do the data show it?
At Station 3 the ball is momentarily at rest near t = 0.5 s. Is the net force zero at that moment? Use the slope to decide.
At Station 4, does the fan thrust account for the whole net force? Compare ma with 0.30 N.
Modification for Distance Learning
Post each station as a slide with its table. Groups work in breakout rooms and submit one card per station on a shared document.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Same Net Force, Different Masses (Velocity-Time)
Invented motion-detector data for two carts, each pulled by a 0.80 N net force, drawn to scale. The slope of each line is the acceleration: 1.6 m/s² for the 0.50 kg cart and 0.8 m/s² for the 1.00 kg cart. Doubling the mass halves the slope, as Fnet/m predicts.
Diagram 2: Acceleration Against Net Force at Constant Mass
The four guided-practice trials for the 0.75 kg cart, drawn to scale. The points lie close to a straight line through the origin, so acceleration is proportional to net force. The slope (about 1.33 per kg) equals 1/m, which supports a = Fnet/m.
04
Homework Assignment
~30 min
HS-PS2-1 Homework: Testing Newton's Second Law with Data
Directions: All data sets are invented for practice, and every motion is along one line. Use g = 9.8 m/s². Show each calculation with units, and end every problem with a one- or two-sentence claim that names your evidence.
Part 1: Acceleration from Data (Problems 1-2)
A 0.25 kg cart is pulled by a force sensor that reads 0.18 N on a low-friction track. Its velocity is 0.20, 0.55, 0.90 and 1.25 m/s at t = 0, 0.5, 1.0 and 1.5 s. (a) Find the acceleration. (b) Compute m × a. (c) By what percent does m × a differ from the sensor reading, and do the data support Newton's second law?
A 3.0 kg box is pulled from rest across a floor by a horizontal rope. Its position is 0, 0.05, 0.20, 0.45 and 0.80 m at t = 0, 0.5, 1.0, 1.5 and 2.0 s. (a) Show that the second differences are equal and find the acceleration. (b) Find the net force on the box. (c) A spring scale in the rope reads 9.0 N. What is the friction force on the box?
Part 2: Graphs of Force, Mass and Acceleration (Problems 3-4)
Carts of different mass are pulled by the same net force. The data are: m = 0.40, 0.80, 1.20, 1.60 kg and a = 2.25, 1.14, 0.74, 0.56 m/s². (a) Compute m × a for each trial. (b) Do the data support the claim that a is proportional to 1/m? (c) Estimate the net force.
A cart of unknown mass is pulled by four different net forces: F = 0.5, 1.0, 1.5, 2.0 N gives a = 0.21, 0.40, 0.61, 0.79 m/s². (a) Graph a against F and draw a best-fit line. (b) Use the slope to find the mass of the cart. (c) Explain why the line should pass through the origin.
Part 3: Vertical Motion and Arguments (Problems 5-6)
A 70 kg student stands on a bathroom scale in an elevator. As the elevator starts upward, its velocity is 0, 0.6, 1.2 and 1.8 m/s at t = 0, 0.5, 1.0 and 1.5 s. (a) Find the acceleration. (b) Find the net force on the student. (c) Predict the scale reading. (d) Predict the reading when the elevator later moves upward at a steady speed.
A student claims: "A heavier ball falls with a larger acceleration because gravity pulls on it harder." Two balls are dropped side by side. The 0.50 kg ball has v = 0, 2.45, 4.90, 7.34 m/s and the 0.10 kg ball has v = 0, 2.44, 4.86, 7.26 m/s at t = 0, 0.25, 0.50, 0.75 s. Use the data and Fnet = ma to write a claim, evidence and reasoning response to the student.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Acceleration from Data
Correct slope or second-difference method with units for every data set
Correct method with one calculation or unit error
Net Force and Comparison
Fnet = ma computed and compared with the measured or predicted force, including friction where it acts
Net force computed but not compared, or friction ignored
Claim, Evidence, Reasoning
Claim names specific data as evidence and explains how it supports Fnet = ma
Claim stated but evidence is general or reasoning is missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All data sets are invented, all motion is along one line, and g = 9.8 m/s². Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
On a velocity-time graph of a cart, which quantity does the slope of the line give?
Answer: B
Slope is change in velocity divided by change in time, which is acceleration. Choice A confuses the graph with a position-time graph. Choice C is only true after multiplying by the mass, and the slope alone cannot give the mass (choice D).
Question 2 of 20 · Multiple Choice
A 2.5 kg cart starts from rest and its velocity rises steadily to 2.4 m/s in 3.0 s. What is the net force on the cart?
Answer: C
The acceleration is 2.4 ÷ 3.0 = 0.8 m/s², so Fnet = ma = 2.5 × 0.8 = 2.0 N. Choice A multiplies the mass by the final velocity instead of the acceleration. Choice B is the acceleration, not the force. Choice D divides the acceleration by the mass.
Question 3 of 20 · Multiple Choice
The net force on a cart is doubled while its mass stays the same. According to the data in this lesson and Newton's second law, what happens to the acceleration?
Answer: D
With the mass fixed, a = Fnet/m is proportional to the net force, so doubling the force doubles the acceleration. Choice B describes doubling the mass instead. Choice C would require a relationship with the square of the force, which the linear acceleration-force graph rules out.
Question 4 of 20 · Multiple Choice
The same net force is applied to a cart and then to a second cart with three times the mass. How does the acceleration of the second cart compare?
Answer: A
a = Fnet/m, so with the force fixed, tripling the mass divides the acceleration by 3. Choice B reverses the relationship. Choice D squares the factor, which the data (m × a stays fixed) do not support.
Question 5 of 20 · Multiple Choice
A cart starts from rest. Its position is 0, 0.25, 1.00 and 2.25 m at t = 0, 1, 2 and 3 s. What is its acceleration?
Answer: C
The second differences are 0.75 - 0.25 = 0.50 m and 1.25 - 0.75 = 0.50 m for 1 s intervals, so a = 0.50 m/s². Also 2x/t² = 2(2.25)/9 = 0.50. Choice A uses x/t² and forgets the factor 2 in x = ½at². Choice B is x/t at 3 s, an average velocity, not an acceleration.
Question 6 of 20 · Multiple Choice
A graph of acceleration against net force for one cart is a straight line through the origin with a slope of 0.25 per kg. What is the mass of the cart?
Answer: B
Newton's second law gives a = (1/m)F, so the slope is 1/m. Then m = 1 ÷ 0.25 = 4.0 kg. Choice A reads the slope itself as the mass, a common error when the axes are not interpreted. Choice C doubles the correct value, and choice D doubles the slope.
Question 7 of 20 · Multiple Choice
A motion detector shows a puck gliding with a velocity of 3.0 m/s that does not change for 4 s. What does Newton's second law say about the net force on the puck during those 4 s?
Answer: A
An unchanging velocity means zero acceleration, and Fnet = ma = 0. Choice B reflects the idea that motion needs a forward force, which the data do not support. Choice D confuses the weight with the net force: the weight is balanced by the support force from the ice.
Question 8 of 20 · Multiple Choice
A 0.50 kg cart moving along a track slows steadily from 1.5 m/s to 0.3 m/s in 1.5 s. What net force acts on it?
Answer: B
The acceleration is (0.3 - 1.5) ÷ 1.5 = -0.8 m/s², so Fnet = 0.50 × (-0.8) = -0.40 N: 0.40 N opposite to the motion. Choice C gives the acceleration value as a force. Choice A misses the sign of the slope.
Question 9 of 20 · Multiple Choice
Four groups pulled the same 0.50 kg cart, and each reported one trial as (net force, measured acceleration). Which result best supports Newton's second law?
Answer: A
Newton's second law predicts a = Fnet/m. For 0.20 N that is 0.20 ÷ 0.50 = 0.40 m/s², and 0.41 m/s² is within about 2.5% of it. For 0.40 N the prediction is 0.80 m/s², so choices C and D are far off. Choice B matches a = F × m, which multiplies by the mass instead of dividing, and choice C matches a = F with the mass ignored.
Question 10 of 20 · Multiple Choice
A 50 kg student stands on a scale in an elevator that is moving upward and slowing down at 0.8 m/s². What does the scale read?
Answer: C
Slowing while moving up means the acceleration points down. Taking up as positive, N - mg = m(-0.8), so N = 50(9.8 - 0.8) = 450 N. Choice B uses the wrong direction for the acceleration. Choice A ignores the acceleration, and choice D is only the net force.
Question 11 of 20 · Multiple Choice
A 3.0 g coffee filter falls and reaches a steady speed of 1.5 m/s. What is the air resistance force on it at that point?
Answer: D
At a steady speed the acceleration is zero, so the net force is zero and the air resistance equals the weight: 0.0030 kg × 9.8 m/s² = 0.0294 N. Choice A gives the net force, not the air resistance. Choice B multiplies the mass by the speed.
Question 12 of 20 · Multiple Choice
A 0.60 kg cart rolls from rest down a ramp. Its velocity is 0, 0.36 and 0.72 m/s at t = 0, 0.4 and 0.8 s. What is the net force along the ramp?
Answer: A
The slope is 0.72 ÷ 0.8 = 0.9 m/s², so Fnet = 0.60 × 0.9 = 0.54 N. Choice B multiplies the mass by the final velocity. Choice C divides the acceleration by the mass. Choice D is the full weight, but only part of the weight acts along the ramp, and friction also acts.
Question 13 of 20 · Multiple Choice
A 1.0 N net force gives a 0.5 kg cart an acceleration of 2.0 m/s² and a 1.0 kg cart an acceleration of 1.0 m/s². If Newton's second law holds, what acceleration should the same force give a 2.0 kg cart?
Answer: D
In both trials m × a = 1.0 N, so for 2.0 kg, a = 1.0 ÷ 2.0 = 0.5 m/s². Choice C multiplies instead of dividing. Choice A halves the acceleration twice, as if a depended on 1/m².
Question 14 of 20 · Multiple Choice
A force sensor and a balance predict Fnet/m = 2.0 m/s² for a cart, but the motion detector measures 1.9 m/s². Which conclusion is best supported?
Answer: C
The difference is (2.0 - 1.9) ÷ 2.0 = 5%, a small gap for classroom equipment. A little friction lowers the net force below the sensor reading, which lowers the acceleration. Choice A rejects the law over a difference that fits normal measurement uncertainty.
Question 15 of 20 · Short Answer
A 2.0 kg cart has velocity 0.4, 1.5, 2.6 and 3.7 m/s at t = 0, 1, 2 and 3 s. Find the acceleration and the net force.
The velocity rises by 1.1 m/s every second, so a = 1.1 m/s² (the starting velocity does not matter). Fnet = 2.0 × 1.1 = 2.2 N in the direction of motion.
Question 16 of 20 · Short Answer
For one cart, net forces of 0.40, 0.80 and 1.20 N give accelerations of 0.50, 1.00 and 1.50 m/s². Do the data support Newton's second law? What is the mass of the cart?
Yes. Each acceleration is exactly proportional to the force (a ÷ F = 1.25 per kg each time), so the a-F graph is a line through the origin. The slope is 1/m, so m = 1 ÷ 1.25 = 0.80 kg.
Question 17 of 20 · Short Answer
After the string goes slack, a 0.50 kg cart has velocity 1.10, 1.09, 1.08 and 1.07 m/s at t = 0, 1, 2 and 3 s. A student says "the cart keeps moving, so a forward net force must still act on it." Use the data to respond.
The slope is (1.07 - 1.10) ÷ 3 = -0.01 m/s², so Fnet = 0.50 × (-0.01) = -0.005 N: a tiny force opposite to the motion (friction), not a forward force. The cart keeps moving because nothing large acts to change its velocity; a net force is needed to change motion, not to keep it.
Question 18 of 20 · Short Answer
A 0.12 kg ball is dropped. Its velocity (downward positive) is 0, 0.49, 0.98 and 1.47 m/s at t = 0, 0.05, 0.10 and 0.15 s. Find the acceleration and the net force, and compare the net force with the ball's weight.
The slope is 0.49 ÷ 0.05 = 9.8 m/s², so Fnet = 0.12 × 9.8 = 1.176 N (about 1.18 N) downward. The weight is mg = 0.12 × 9.8 = 1.176 N, so the net force equals the weight: air resistance is negligible over this short fall.
Question 19 of 20 · Short Answer
You want to use data to test whether acceleration is proportional to 1/m when the net force is fixed. What quantities do you measure, what graph do you draw, and what shape supports the claim?
Keep the net force fixed (for example, the same hanging mass or the same force sensor reading) and change the mass of the cart. Measure the mass on a balance and find the acceleration from velocity-time or position-time data. Plot a against 1/m: a straight line through the origin, with slope equal to the net force, supports a = Fnet/m. A graph of a against m would be a curve, which is harder to judge.
Question 20 of 20 · Short Answer
A 25 kg sled is pulled across level snow by a horizontal rope with a force of 60 N. Its velocity is 0, 0.8, 1.6 and 2.4 m/s at t = 0, 1, 2 and 3 s. Find the net force and the friction force.
The slope is a = 0.8 m/s², so Fnet = 25 × 0.8 = 20 N. The net force is the rope force minus friction: 60 - f = 20, so friction is 40 N, opposite to the motion.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-PS2-1 mean?
HS-PS2-1 asks students to use motion data to show that net force, mass and acceleration are related by Newton's second law, Fnet = ma. The focus is on analyzing data, not only on memorizing the formula: students find acceleration from tables or graphs and check it against Fnet/m. The code means High School, Physical Science, core idea 2 (Motion and Stability: Forces and Interactions), performance expectation 1.
Is HS-PS2-1 taught in physics or physical science?
It is usually taught in a high school Physics course, and it can also sit in Physical Science or in the physics unit of an integrated science course. The math needed is slope, proportional reasoning and rearranging F = ma, so it fits students who have finished or are taking Algebra I.
What is the assessment boundary for HS-PS2-1?
The official Assessment Boundary limits assessment to one-dimensional motion of macroscopic objects at non-relativistic speeds. Forces and motion along a single line (a level track, a ramp, straight up and down) are in scope. Two-dimensional motion such as projectiles, circular motion, and objects near the speed of light are beyond this standard.
What kind of data do students analyze for Newton's second law?
The Clarification Statement names tables or graphs of position or velocity against time for objects with a net unbalanced force, such as a falling object, an object rolling down a ramp, or an object pulled by a constant force. Classroom sources include motion detectors, phone video stepped frame by frame, photogates, or a stopwatch and meter stick. Invented data sets like the ones on this page work well for practice and for students who miss a lab.
How do you find acceleration from a velocity-time graph?
Find the slope: choose two points far apart on the best-fit line, subtract the velocities and divide by the time between them. For example, if the velocity goes from 0.5 m/s to 3.5 m/s over 4.0 s, the acceleration is 3.0 ÷ 4.0 = 0.75 m/s². Use the line rather than two data points when the data are noisy.
How do you get acceleration from position-time data?
If the object starts from rest and the acceleration is steady, use x = ½at², so a = 2x/t². With equal time steps, you can also take the second differences of the positions and divide by the time step squared. A position-time graph for steady acceleration is a curve (half of a parabola), not a straight line.
Why doesn't lab data match F = ma exactly?
Real data always carry some error. Friction and air resistance make the net force smaller than the applied force, timing by hand adds reaction-time error, and strings, pulleys and track tilt add small forces. A gap of a few percent is normal for classroom equipment. Ask students to name a source of error and say whether it would make the measured acceleration larger or smaller than Fnet/m.
What is the difference between applied force and net force?
The applied force is one push or pull; the net force is the sum of all forces along the line of motion, with opposite directions given opposite signs. Newton's second law uses the net force. If a rope pulls a crate with 45 N and friction acts with 30 N in the other direction, the net force is 15 N.
Can students do HS-PS2-1 without force sensors or motion detectors?
Yes. A cart pulled by a hanging mass over a pulley gives a known force (the weight of the hanging mass), and timing the cart over a measured distance from rest gives a = 2d/t². Phone video with a meter stick in the frame gives position-time data for falling or rolling objects. The practice of analyzing data is the same with low-cost equipment.
What comes after HS-PS2-1?
HS-PS2-2 uses mathematical representations to show that total momentum is conserved when no net force acts on a system, and HS-PS2-3 applies force ideas to design a device that reduces the force in a collision. Both build on the relationship among force, mass and change in motion from HS-PS2-1.
07
Related Standards
5 standards
These standards connect to HS-PS2-1: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-PS2-2Prerequisite
Plan an investigation showing that change in motion depends on net force and mass
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