C. Forces · ForcesInvestigating Acceleration, Net Force and Mass
C2.3 C2.2 A1.12About 70 minutes
Learning goals- I can describe an investigation of how acceleration depends on net force and on mass, and identify its variables.
- I can graph and analyse data to show that a ∝ Fnet and a ∝ 1/m.
- I can use the slope of a best-fit line to find a mass or a force, and calculate a percent error.
- I can explain why real data do not fit the theory perfectly.
Words to know
independent variable
The variable the experimenter changes on purpose (here, the net force or the mass).
dependent variable
The variable that is measured to see how it responds (here, the acceleration).
controlled variables
Variables kept the same so they do not affect the result.
line of best fit
A straight line drawn through data points so that the points are scattered evenly on both sides of it.
slope
rise ÷ run of a line on a graph, found from two points on the best-fit line (not from data points).
percent error
|experimental value − accepted value| ÷ accepted value × 100%.
Test the law, don't just trust it
Newton's second law is a claim about nature. In science, a claim earns our trust only when experiments support it. In this lesson you will look at how physics classes test Fnet = ma, analyse real-looking data the way a scientist does, and learn the graphing skills you will use in every science course from here on, including SPH4U.
The set-up
A low-friction cart runs on a level track. A string runs from the cart over a pulley to a hanging mass. A force sensor on the cart measures the pull of the string; a motion sensor measures the cart's velocity many times per second, and software finds the acceleration from the slope of the v-t graph (as in the Kinematics unit).
Investigation 1: change the force, keep the mass constant.
- Independent variable: net force on the cart.
- Dependent variable: acceleration.
- Controlled: mass of the cart, track, level surface.
Investigation 2: change the mass, keep the force constant.
- Independent variable: mass of the cart (add bars of known mass).
- Dependent variable: acceleration.
- Controlled: the pulling force.
A small trick: friction on a cart is small but not zero. Some teachers tilt the track very slightly, just enough that a gentle tap makes the cart roll at a steady speed. Then gravity along the track balances friction, and the string's pull is the net force.
| Force F (N) | 0.60 | 1.20 | 1.80 | 2.40 | 3.00 |
|---|
| Acceleration a (m/s²) | 0.46 | 0.97 | 1.47 | 1.98 | 2.46 |
Worked exampleExample 1: Analysing a against F
Use the Investigation 1 data. (a) What does the graph show? (b) Find the slope of the best-fit line and use it to find the cart's mass. (c) Calculate the percent error compared with the measured mass, 1.2 kg.
- (a) Plot a (dependent) on the vertical axis and F (independent) on the horizontal axis. The points lie very close to a straight line almost through the origin, so a is directly proportional to F.
- (b) Pick two points far apart on the best-fit line (not data points): about (0.60 N, 0.46 m/s²) and (3.00 N, 2.46 m/s²). Slope = (2.46 − 0.46) ÷ (3.00 − 0.60) = 2.00 ÷ 2.40 ≈ 0.83 m/s² per N (= 0.83 /kg).
- From a = (1/m)F, the slope is 1/m. So m = 1 ÷ 0.833 ≈ 1.2 kg.
- (c) Percent error = |1.20 − 1.2| ÷ 1.2 × 100% ≈ 0% to two significant figures. The line meets the F-axis at about 0.05 N: that small force is needed just to overcome friction.
a ∝ F; slope ≈ 0.83 /kg, so m ≈ 1.2 kg, matching the measured mass.
Acceleration against net force for a 1.2 kg cart. The points lie close to a straight line, so a is directly proportional to F. The slope is 1/m.| Mass m (kg) | 0.50 | 0.75 | 1.00 | 1.50 | 2.00 |
|---|
| 1/m (1/kg) | 2.00 | 1.33 | 1.00 | 0.67 | 0.50 |
| Acceleration a (m/s²) | 2.9 | 2.0 | 1.5 | 1.0 | 0.76 |
Worked exampleExample 2: Straightening a curve
Use the Investigation 2 data. (a) Describe the graph of a against m. (b) Why do scientists graph a against 1/m instead? (c) Find the slope of a against 1/m and the percent error compared with 1.5 N.
- (a) As m increases, a decreases, quickly at first and then more slowly: a curve. Doubling m from 0.50 to 1.00 kg roughly halves a (2.9 to 1.5); doubling again to 2.00 kg roughly halves it again (to 0.76).
- (b) It is hard to tell exactly which curve fits. If a = F/m, then graphing a against 1/m should give a straight line through the origin, which is easy to test and whose slope is F.
- (c) The points lie close to a straight line. Using the best-fit line at 1/m = 0.50 (a ≈ 0.76) and 1/m = 2.00 (a ≈ 2.9): slope = (2.9 − 0.76) ÷ (2.00 − 0.50) = 2.14 ÷ 1.50 ≈ 1.4 N.
- Percent error = |1.43 − 1.5| ÷ 1.5 × 100% ≈ 5%. The slope is a little low, probably because some friction acted on the cart, so the net force was a little less than 1.5 N.
a ∝ 1/m; slope ≈ 1.4 N, about 5% below the expected 1.5 N.
Graphing a against 1/m (not m) turns the curve into a straight line through the origin, showing that a is inversely proportional to m. The slope is the net force.When the net force changes during the motion
In these labs the net force stays constant, so the acceleration does too. In real life, air resistance grows with speed. A falling object's net force (gravity minus air resistance) gets smaller as it speeds up, so its acceleration gets smaller, until the net force is zero and it falls at a constant terminal velocity. Computer simulations let you change the mass and shape and watch this happen. You will study it in the free-fall lesson of this unit.
Sources of error
- Friction in the wheels and pulley (makes measured a too small).
- Mass of the string and the pulley's own inertia.
- Track not level (gravity helps or hinders).
- Sensor reading errors; choosing the wrong part of the v-t graph.
Good practice: repeat each trial (three or more), average, and draw best-fit lines instead of joining the dots.
Just for funLab-cart tip: a graph that goes perfectly through every point is not a sign of great physics. It is a sign someone "fixed" the data. Real data wiggle a little, like a cart with one sticky wheel.
Hands-on activityHome investigation: rubber-band launcher
You need: a toy car or a small box on smooth floor, a thick rubber band, two chairs (or a doorway), a tape measure, tape, and coins or small weights you can tape to the car.
- Loop the rubber band between two chair legs to make a slingshot about 1 cm off the floor.
- Pull the car back by a fixed amount (mark it, e.g. 10 cm) and release. Measure how far it travels. Repeat 3 times and average.
- Keep the pull-back the same (same force) and tape on extra mass: +50 g, +100 g, +150 g. Measure the distance each time.
- Then keep the mass fixed and change the pull-back: 5 cm, 10 cm, 15 cm.
- Graph distance against added mass, and distance against pull-back. Describe the patterns.
- Explain why distance is only a rough stand-in for acceleration here (the band's force changes as it un-stretches, and friction slows the car afterward).
Aim the launcher at the floor, away from people and pets, and wear eye protection or glasses if you have them; a snapping rubber band stings.
Your turn
1In Investigation 1, name the independent, dependent and two controlled variables.
Independent: net force. Dependent: acceleration. Controlled: cart mass, same track and level surface (also the same sensor settings).
2A best-fit line of a against F has a slope of 0.50 /kg. What is the mass of the cart?
Slope = 1/m, so m = 1 ÷ 0.50 = 2.0 kg.
3Why do we use two points on the best-fit line, not two data points, to find a slope?
Data points contain random error. The best-fit line averages out the scatter of all the points, so its slope is more reliable.
4In Investigation 2 a student graphs a against m and gets a curve. What should she graph to get a straight line, and what does its slope mean?
Graph a against 1/m. The slope is the net force.
5Measured value 9.5, accepted value 10.0. Find the percent error.
|9.5 − 10.0| ÷ 10.0 × 100% = 5.0%.
6Give two reasons a cart's measured acceleration might be less than F/m predicts.
Any two: friction in wheels or pulley; mass of string or pulley not counted; track tilted slightly uphill; sensor error.
More practice: warm-up
1Which variable goes on the horizontal axis?
The independent variable.
2What shape is the a-F graph when mass is constant?
A straight line through (or very near) the origin.
3What happens to a when m doubles and F stays the same?
4What does the motion sensor measure?
The cart's position and velocity over time (from which acceleration is found).
5Why repeat each trial several times?
To reduce the effect of random errors by averaging.
More practice: on your own
1A cart is pulled with forces of 1.0, 2.0, 3.0 and 4.0 N; the accelerations are 0.40, 0.80, 1.20 and 1.60 m/s². Find the cart's mass.
Slope = 1.60 ÷ 4.0 = 0.40 /kg, so m = 1 ÷ 0.40 = 2.5 kg. (Each a ÷ F = 0.40.)
2In a lab, the best-fit line of a against F crosses the F-axis at 0.20 N instead of at 0. Explain what this means.
Forces below about 0.20 N did not accelerate the cart: that much force was needed just to balance friction. The net force is really F − 0.20 N.
3With F = 2.0 N, a 0.80 kg cart should accelerate at what rate? A student measures 2.3 m/s². Find the percent error.
a = 2.0 ÷ 0.80 = 2.5 m/s². Error = |2.3 − 2.5| ÷ 2.5 × 100% = 8.0%.
4Why do some teachers tilt the track slightly before the experiment?
So that a component of gravity along the track balances friction: the cart rolls at a steady speed with no string. Then the string's pull equals the net force.
5In a lab using hanging masses, a student moves 50 g masses from the cart to the hanger one at a time. Why is this better than just adding masses to the hanger?
The hanging mass is part of the system that accelerates. Moving masses from the cart to the hanger increases the pulling force while keeping the total mass constant, so only one variable changes.
Challenge
1Data for a cart: m = 0.50, 1.0, 2.0 kg with a = 3.8, 2.0, 1.0 m/s² (same force). Calculate 1/m, find the slope of a against 1/m from the first and last points, and state the force.
1/m = 2.0, 1.0, 0.50 /kg. Slope = (3.8 − 1.0) ÷ (2.0 − 0.50) = 2.8 ÷ 1.5 ≈ 1.9 N. The force is about 1.9 N (the middle point, 2.0 × 1.0 = 2.0 N, agrees closely).
2A student's a-F graph is a straight line but its slope gives a mass of 1.5 kg while the cart's mass is 1.2 kg. Suggest a reason, assuming the student pulled with hanging masses and used F = (hanging mass) × g.
The hanging mass also accelerates, so the total mass being accelerated is cart + hanging mass, not just the cart. The slope gives 1/(total mass). Also, F = mg for the hanging mass is the pull only if it is not accelerating; the actual string tension is smaller. Either way the graph reflects more than the cart's mass.
3Explain how you could use a computer simulation of a falling object with air resistance to show that a decreases as speed increases, even though the mass is constant.
Record a and v at several times. As v increases, air resistance grows, so Fnet = mg − Fair decreases; the simulation shows a dropping toward zero, while m stays fixed. Plotting a against Fnet should still give a straight line with slope 1/m, showing the second law holds at every instant.
Quiz
1. In an experiment where the mass is kept constant and the force is changed, the dependent variable is:
2. A graph of a against Fnet for a cart is a straight line through the origin. This shows that:
3. The slope of a graph of a against Fnet equals:
4. To get a straight line when the force is constant and the mass changes, plot:
5. Expected value 4.0 m/s², measured value 3.6 m/s². The percent error is:
6. Which is a likely reason measured accelerations are slightly smaller than predicted?
Big idea: Students investigate and analyse data showing how acceleration depends on net force and mass (C2.3), using graphs, slopes and percent error (A1.12).
How to help: Ask your teen to explain the graphs: "What's on each axis? Why is it a straight line? What does the slope tell you?" Encourage neat graphs with labelled axes, units, a title, and a ruler-drawn best-fit line.
Common difficulties:
- Joining the dots instead of drawing a best-fit line.
- Using data points rather than points on the line for the slope.
- Forgetting that the slope of a-F is 1/m, not m.
- Treating any difference from the theory as "wrong". Explaining sources of error is part of good science.