Respuesta :
Spring force is given be kx.
Where k is spring constant and x is the distance moved.
Using this relation,
[tex]5N = k(2 \times {10}^{ - 2}m ) \\ \\ k = 250N {m}^{ - 1} [/tex]
Where k is spring constant and x is the distance moved.
Using this relation,
[tex]5N = k(2 \times {10}^{ - 2}m ) \\ \\ k = 250N {m}^{ - 1} [/tex]
Answer:
Use Hooke's law
Explanation:
- Apply formula: F = kx
- Sub values in: 5 = 0.02k
- Use algebra: k = 250
- The spring constant is 250 N/m