Newton's second law: calculations and data

Answer Key - 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.

   1.96 m/s²
2. 2.

   Bo's 1.5 kg cart on a level track is pulled by a string over a pulley from a 0.5 kg hanging mass. Ignoring friction, what is the cart's acceleration? Use g = 9.8 m/s².

   2.45 m/s²
3. 3.

   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.

   3.92 m/s²

4. 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

5. 4.

   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 0.5 kg from the slope, friction 1.0 N from the intercept
   • Mass 2.0 kg and friction 0 N, since the line is straight
   • Mass 2.0 kg from the slope, friction 2.0 N, the first force reading
6. 5.

   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?

   • 6.53 m/s²: 19.6 N ÷ 3 kg
   • 7.35 m/s²: (19.6 − 4.9) N ÷ 2 kg
   • 4.9 m/s²: (19.6 − 4.9) N ÷ 3 kg
   • 3.27 m/s²: (19.6 − 9.8) N ÷ 3 kg

7. 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

8. 6.

   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
   • Friction, because it takes the same 0.04 m/s² in each row
   • Random error, because the measured values scatter around the predicted ones
   • F = ma fails, because no row matches the prediction

9. 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

10. 7.

    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
11. 8.

    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
    • No: the mass is 0.4² = 0.16 kg
    • Yes: the slope of an acceleration-force graph is the mass
    • No: the slope is 1 ÷ mass, so the mass is 2.5 kg
12. 9.

    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
    • 12 N: 2 kg × 6 m/s², from 30 N ÷ 5 kg
    • 12.4 N: T − 2 N = 2 kg × 5.2 m/s²
    • 10.4 N: 2 kg × 5.2 m/s²

13. 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

14. 10.

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

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