11th grade Newton's second law: calculations and data worksheet

Newton's second law: calculations and data

Grade 11 Science Worksheet - Challenge Me

Circle the best answer, or write your answer on the line.

1. 1.

   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.
2. 2.

   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.

3. 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

4. 3.

   Jun plots applied force against acceleration for one cart. What do the slope and the intercept give?

   • 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 2.0 kg and friction 0 N, since the line is straight
   • Mass 0.5 kg from the slope, friction 1.0 N from the intercept

5. 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

6. 4.

   Luz predicts a = mg ÷ 1.00 kg for the hanging-mass lab. Which conclusion does the table support?

   • A steady ratio of 1.09 shows about 0.09 kg extra is moving, such as the pulley
   • F = ma fails, because no row matches the prediction
   • Friction, because it takes the same 0.04 m/s² in each row
   • Random error, because the measured values scatter around the predicted ones

7. 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

8. 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: the mass changed between the rows, which explains the gap
   • No: F − 2 N equals 2 kg × a in each row; 2 N of friction was omitted
   • No: friction makes the acceleration larger than F ÷ m, so the law still holds

9. Two trials (example data)

   trial	mass (kg)	net force (N)	acceleration (m/s²)
   1	2	6	3
   2	4	10	2.5

10. 6.

    Noah says doubling the mass should halve the acceleration, so trial 2 breaks the law. Which reply fits the table?

    • Yes: 3 and 2.5 are close, so the law is approximate in practice
    • No: mass has no effect on acceleration
    • No: the force also changed, 6 N to 10 N, and F ÷ m fits both
    • Yes: doubling the mass should give 1.5 m/s², but trial 2 shows 2.5
11. 7.

    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?

    • No: the mass is 0.4 × 9.8 = 3.9 kg
    • 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
12. 8.

    A 30 N pull on a 3 kg cart tows a 2 kg cart by a light rope. Each cart has 2 N of friction. What is the rope tension, Ren?

    • 14.4 N: 2 kg × 5.2 m/s² + 4 N
    • 10.4 N: 2 kg × 5.2 m/s²
    • 12 N: 2 kg × 6 m/s², from 30 N ÷ 5 kg
    • 12.4 N: T − 2 N = 2 kg × 5.2 m/s²
13. 9.

    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?

    • 1.65 to 2.47 m/s²; yes, 2.0 and 2.3 are clearly different
    • 1.65 to 2.47 m/s²; no, 2.0 and 2.3 both fit
    • 1.9 to 2.1 m/s²; yes, a 5% spread is small
    • 1.8 to 2.2 m/s²; yes, one run separates 2.0 from 2.3

14. 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

15. 10.

    Zee tests a 2.0 kg cart, comparing each measured a with F ÷ m. What do the differences show?

    • Random scatter: the differences change sign, so no systematic error
    • F = ma fails, because no row matches exactly
    • A systematic error: the differences grow with the force
    • A steady error: each row differs from F ÷ m by the same amount