Newton's second law: calculations and data
Grade 11 Science Worksheet - Challenge Me
Circle the best answer, or write your answer on the line.
Ari's 4 kg cart on a level track is pulled by a string over a pulley from a 1 kg hanging mass. Ignoring friction, what is the cart's acceleration? Use g = 9.8 m/s².
Esme's 3 kg cart on a level track is pulled by a string over a pulley from a 2 kg hanging mass. Ignoring friction, what is the cart's acceleration? Use g = 9.8 m/s².
Read this first
Force against acceleration for one cart (example data)
acceleration (m/s²) applied force (N) 0.5 2 1 3 1.5 4 2 5 Jun plots applied force against acceleration for one cart. What do the slope and the intercept give?
- Mass 2.0 kg and friction 0 N, since the line is straight
- The slope is the mass, 2.0 kg; the intercept is friction, 1.0 N
- Mass 2.0 kg from the slope, friction 2.0 N, the first force reading
- Mass 0.5 kg from the slope, friction 1.0 N from the intercept
A 1 kg block on a smooth ramp is tied over a pulley to a 2 kg hanging mass. Down the ramp, gravity pulls the block with 4.9 N. What is the acceleration, Kira?
- 4.9 m/s²: (19.6 − 4.9) N ÷ 3 kg
- 3.27 m/s²: (19.6 − 9.8) N ÷ 3 kg
- 6.53 m/s²: 19.6 N ÷ 3 kg
- 7.35 m/s²: (19.6 − 4.9) N ÷ 2 kg
Hanging mass lab, predicted and measured (example data)
hanging mass (kg) measured a (m/s²) predicted a (m/s²) 0.05 0.45 0.49 0.1 0.9 0.98 0.15 1.35 1.47 0.2 1.8 1.96 Luz predicts a = mg ÷ 1.00 kg for the hanging-mass lab. Which conclusion does the table support?
- Random error, because the measured values scatter around the predicted ones
- A steady ratio of 1.09 shows about 0.09 kg extra is moving, such as the pulley
- Friction, because it takes the same 0.04 m/s² in each row
- F = ma fails, because no row matches the prediction
A 2 kg cart pulled with different forces (example data)
applied force (N) measured acceleration (m/s²) 3 0.5 5 1.5 7 2.5 9 3.5 Milo says the table breaks F = ma, since 3 N on a 2 kg cart should give 1.5 m/s², not 0.5. Which reply fits the data?
- Yes: F ÷ m predicts 1.5, 2.5, 3.5 and 4.5 m/s², and the measured rows do not fit
- No: F − 2 N equals 2 kg × a in each row; 2 N of friction was omitted
- No: the mass changed between the rows, which explains the gap
- No: friction makes the acceleration larger than F ÷ m, so the law still holds
Two trials (example data)
trial mass (kg) net force (N) acceleration (m/s²) 1 2 6 3 2 4 10 2.5 Noah says doubling the mass should halve the acceleration, so trial 2 breaks the law. Which reply fits the table?
- No: mass has no effect on acceleration
- Yes: doubling the mass should give 1.5 m/s², but trial 2 shows 2.5
- No: the force also changed, 6 N to 10 N, and F ÷ m fits both
- Yes: 3 and 2.5 are close, so the law is approximate in practice
Pia reads a slope of 0.4 on the acceleration-force graph and says the cart's mass is 0.4 kg. Which reply is right?
- Yes: the slope of an acceleration-force graph is the mass
- No: the mass is 0.4² = 0.16 kg
- No: the slope is 1 ÷ mass, so the mass is 2.5 kg
- No: the mass is 0.4 × 9.8 = 3.9 kg
A cart starts from rest and covers 1.00 m in a hand-timed 1.0 ± 0.1 s. What range of acceleration fits, and can one run tell 2.0 from 2.3 m/s², Tala?
- 1.8 to 2.2 m/s²; yes, one run separates 2.0 from 2.3
- 1.9 to 2.1 m/s²; yes, a 5% spread is small
- 1.65 to 2.47 m/s²; no, 2.0 and 2.3 both fit
- 1.65 to 2.47 m/s²; yes, 2.0 and 2.3 are clearly different
A 2.0 kg cart, measured against F divided by m (example data)
net force (N) measured a (m/s²) 4 1.9 8 4.1 12 5.8 16 8.2 Zee tests a 2.0 kg cart, comparing each measured a with F ÷ m. What do the differences show?
- A steady error: each row differs from F ÷ m by the same amount
- A systematic error: the differences grow with the force
- Random scatter: the differences change sign, so no systematic error
- F = ma fails, because no row matches exactly