Newton's second law: calculations and data

Answer Key - Grade 11 Science Worksheet

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

1. 1.

   A cart with friction has an acceleration-applied force line through (2.5 N, 0.5 m/s²) and (6.5 N, 2.5 m/s²). What is its mass, Ari?

   2 kg
2. 2.

   A cart with friction has an acceleration-applied force line through (5 N, 0.5 m/s²) and (9 N, 1.5 m/s²). What is its mass, Bo?

   4 kg
3. 3.

   A cart with friction has an acceleration-applied force line through (4.5 N, 0.5 m/s²) and (10.5 N, 1.5 m/s²). What is its mass, Esme?

   6 kg

4. Read this first

   A cart pulled with different forces (example data)

   applied force (N)	acceleration (m/s²)
   2	0.3
   4	1.3
   6	2.3
   8	3.3

5. 4.

   Jun fits a line to the table. What mass and friction does the fit give?

   • Mass 6.7 kg and friction 1.4 N, using F ÷ a at the first row
   • Mass 2.0 kg and friction 0.3 N, reading the first a as friction
   • Mass 2.0 kg and friction 1.4 N: slope 0.5, crossing at 1.4 N
   • Mass 0.5 kg and friction 1.4 N, reading the slope as mass
6. 5.

   A cart of mass 2.0 kg gives the table. Luz compares each measured a with F ÷ m. What does the pattern show?

   • F = ma is false for this cart, since no row matches F ÷ m
   • A steady 0.7 m/s² shortfall points to 1.4 N of uncounted friction
   • The cart is lighter than 2.0 kg, since each a is lower
   • Random timing error: some readings are high and some low
8. 6.

   Milo reads the graph. What does the crossing at 1 N on the force axis show?

   • The cart weighs 1 N, as shown at the crossing
   • The cart's mass is 1 kg, read from the crossing
   • The cart already speeds up even with no pull applied
   • Friction of 1 N: the pull must beat it first
9. 7.

   Noah reads the graph. What is the cart's mass?

   Use the picture in question 6.

   • 2.5 kg: 5 N ÷ 2 m/s², ignoring the friction
   • 1 kg: where the line meets the force axis
   • 2 kg: 1 ÷ the slope of 0.5
   • 0.5 kg: the value of the slope itself, read off the graph
10. 8.

    Pia reads the graph. What acceleration does a 4 N applied force give?

    Use the picture in question 6.

    • 1.2 m/s²: (4 − 1) N ÷ 2.5 kg
    • 1.5 m/s²: (4 − 1) N ÷ 2 kg
    • 2 m/s²: 4 N ÷ 2 kg, leaving friction out
    • 2.5 m/s²: (4 + 1) N ÷ 2 kg
11. 9.

    A 30 N pull on a 3 kg cart tows a 2 kg cart behind it by a light rope on a smooth floor. What is the rope tension, Ren?

    • 30 N, the whole pull
    • 12 N: the rope alone accelerates the 2 kg cart at 6 m/s²
    • 15 N, half of the pull
    • 18 N: 3 kg × 6 m/s², the front cart's net force
12. 10.

    A 3 kg mass and a 2 kg mass hang on a light string over a frictionless pulley and are released. What is their acceleration, Tala? Use g = 9.8 m/s².

    • 0.98 m/s²: (3 − 2) × 9.8 ÷ 10
    • 5.88 m/s²: 3 × 9.8 ÷ 5
    • 4.9 m/s²: (3 − 2) × 9.8 ÷ 2
    • 1.96 m/s²: (3 − 2) × 9.8 ÷ (3 + 2)