1. F = 0.25N
2. k = 80 N/m
3. x = 0.2m
4. a = 1m/s²
5. x = 42.6m
Explanation:
1. F = kx
 F = 2.5 X 0.1
 F = 0.25N
2. x = 0.25m
  F = 20 N
  F = kx
  20 = k X 0.25
  k = 80 N/m
3. k = 3500 N/m
  F = 700 N
  x = ?
  700 = 3500 x
  x = 0.2m
4. m = 20kg
  k = 244 N/m
  x = 0.1m
  F = 20N
  a = ?
  F = ma
  20 = 20 a
  a = 1m/s²
5. k = 2.3 N/m
  m = 10kg
  x = ?
  F = kx
  10 X 9.8 = 2.3 x
  x = 42.6m
The solutions of the given problems are required.
The force is 0.25 N
The spring constant is 80 N/m
The displacement is 0.2 m.
The acceleration is [tex]0.22\ \text{m/s}^2[/tex]
The displacement is 42.65 m.
F = Force
m = Mass
k = Spring constant
x = Displacement of spring
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
x = 0.1 m
k = 2.5 N/m
Force is given by
[tex]F=kx\\\Rightarrow F=2.5\times 0.1\\\Rightarrow F=0.25\ \text{N}[/tex]
F = 20 N
x = 0.25 m
[tex]k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{20}{0.25}\\\Rightarrow k=80\ \text{N/m}[/tex]
k = 3500 N/m
W = 700 N
The force balance is
[tex]W=kx\\\Rightarrow x=\dfrac{W}{k}\\\Rightarrow x=\dfrac{700}{3500}\\\Rightarrow x=0.2\ \text{m}[/tex]
m = 20 kg
k = 244 N/m
x = 0.1 m
[tex]F_a[/tex] = Applied force = 20N
Force balance is
[tex]F_{net}=F-F_a\\\Rightarrow ma=kx-F_a\\\Rightarrow a=\dfrac{kx-F_a}{m}\\\Rightarrow a=\dfrac{244\times 0.1-20}{20}\\\Rightarrow a=0.22\ \text{m/s}^2[/tex]
k = 2.3 N/m
m = 10 kg
Force balance is
[tex]mg=kx\\\Rightarrow x=\dfrac{mg}{k}\\\Rightarrow x=\dfrac{10\times 9.81}{2.3}\\\Rightarrow x=42.65\ \text{m}[/tex]
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