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Hooke's Law worksheet

Grades 9–10 · CCSS HSA.CED.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Solve for: Spring constant
- Applied force (F): 42 N
- Displacement from rest length (x): 16 cm
- Show spring constant in: Newtons per meter (N/m)
Find: Spring constant
Answer
2.
Problem 2.
- Solve for: Spring constant
- Applied force (F): 23 N
- Displacement from rest length (x): 12 cm
- Show spring constant in: Newtons per millimeter (N/mm)
Find: Spring constant
Answer
3.
Problem 3.
- Solve for: Force
- Spring constant (k): 180 N/m
- Displacement from rest length (x): 13 cm
- Show force in: Newtons (N)
Find: Applied force
Answer
4.
Problem 4.
- Solve for: Displacement
- Applied force (F): 36 N
- Spring constant (k): 180 N/m
- Show displacement in: Meters (m)
Find: Displacement (extension)
Answer
5.
Problem 5.
- Solve for: Spring constant
- Applied force (F): 30 N
- Displacement from rest length (x): 12 cm
- Show spring constant in: Newtons per centimeter (N/cm)
Find: Spring constant
Answer
6.
Problem 6.
- Solve for: Force
- Spring constant (k): 200 N/m
- Displacement from rest length (x): 22 cm
- Show force in: Pounds-force (lbf)
Find: Applied force
Answer
7.
Problem 7.
- Solve for: Displacement
- Applied force (F): 26 N
- Spring constant (k): 275 N/m
- Show displacement in: Feet (ft)
Find: Displacement (extension)
Answer
8.
Problem 8.
- Solve for: Force
- Spring constant (k): 240 N/m
- Displacement from rest length (x): 12 cm
- Show force in: Pounds-force (lbf)
Find: Applied force
Answer
9.
Problem 9.
- Solve for: Spring constant
- Applied force (F): 30 N
- Displacement from rest length (x): 11 cm
- Show spring constant in: Newtons per millimeter (N/mm)
Find: Spring constant
Answer
10.
Problem 10.
- Solve for: Force
- Spring constant (k): 200 N/m
- Displacement from rest length (x): 10 cm
- Show force in: Newtons (N)
Find: Applied force
Answer

Answer key: Hooke's Law worksheet

1. Spring constant: 262.5 N/m
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $16 cm \times 0.01$ $= 0.16 m$
2. To find the stiffness k, divide the force F by the stretch x
  $\frac{42}{0.16}$ $= 262.5 N/m$
3. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 262.5 \times {0.16}^{2}$ $= 3.36 J$
2. Spring constant: 0.19 N/mm
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $12 cm \times 0.01$ $= 0.12 m$
2. To find the stiffness k, divide the force F by the stretch x
  $\frac{23}{0.12}$ $= 191.666667 N/m$
3. Change the answer to N/mm, the unit you picked: divide by 1,000
  $\frac{191.666667}{1,000}$ $= 0.191667 N/mm$
4. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 191.666667 \times {0.12}^{2}$ $= 1.38 J$
3. Applied force: 23.4 N
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $13 cm \times 0.01$ $= 0.13 m$
2. By Hooke's law, multiply the stiffness k by the stretch x to get the force F
  $180 \times 0.13$ $= 23.4 N$
3. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 180 \times {0.13}^{2}$ $= 1.521 J$
4. Displacement (extension): 0.2 m
1. To find the stretch x, divide the force F by the stiffness k
  $\frac{36}{180}$ $= 0.2 m$
2. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 180 \times {0.2}^{2}$ $= 3.6 J$
5. Spring constant: 2.5 N/cm
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $12 cm \times 0.01$ $= 0.12 m$
2. To find the stiffness k, divide the force F by the stretch x
  $\frac{30}{0.12}$ $= 250 N/m$
3. Change the answer to N/cm, the unit you picked: divide by 100
  $\frac{250}{100}$ $= 2.5 N/cm$
4. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 250 \times {0.12}^{2}$ $= 1.8 J$
6. Applied force: 9.89 lbf
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $22 cm \times 0.01$ $= 0.22 m$
2. By Hooke's law, multiply the stiffness k by the stretch x to get the force F
  $200 \times 0.22$ $= 44 N$
3. Change the answer to lbf, the unit you picked: divide by 4.448221615261
  $\frac{44}{4.448221615261}$ $= 9.891593 lbf$
4. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 200 \times {0.22}^{2}$ $= 4.84 J$
7. Displacement (extension): 0.31 ft
1. To find the stretch x, divide the force F by the stiffness k
  $\frac{26}{275}$ $= 0.094545 m$
2. Change the answer to ft, the unit you picked: divide by 0.3048
  $\frac{0.094545}{0.3048}$ $= 0.310188 ft$
3. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 275 \times {0.094545}^{2}$ $= 1.229091 J$
8. Applied force: 6.47 lbf
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $12 cm \times 0.01$ $= 0.12 m$
2. By Hooke's law, multiply the stiffness k by the stretch x to get the force F
  $240 \times 0.12$ $= 28.8 N$
3. Change the answer to lbf, the unit you picked: divide by 4.448221615261
  $\frac{28.8}{4.448221615261}$ $= 6.474498 lbf$
4. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 240 \times {0.12}^{2}$ $= 1.728 J$
9. Spring constant: 0.27 N/mm
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $11 cm \times 0.01$ $= 0.11 m$
2. To find the stiffness k, divide the force F by the stretch x
  $\frac{30}{0.11}$ $= 272.727273 N/m$
3. Change the answer to N/mm, the unit you picked: divide by 1,000
  $\frac{272.727273}{1,000}$ $= 0.272727 N/mm$
4. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 272.727273 \times {0.11}^{2}$ $= 1.65 J$
10. Applied force: 20 N
1. Change displacement to m, the standard unit the formula uses (1 cm = 0.01 m)
  $10 cm \times 0.01$ $= 0.1 m$
2. By Hooke's law, multiply the stiffness k by the stretch x to get the force F
  $200 \times 0.1$ $= 20 N$
3. For the energy stored in the spring, take half of k times the stretch squared
  $0.5 \times 200 \times {0.1}^{2}$ $= 1 J$