A tiny spring, with a spring constant of 1.20 N/m, will be stretched to what displacement by a 0.005 0-N force?
f=kx
x= f/k
To find the displacement of the tiny spring when a force of 0.0050 N is applied, we can use Hooke's Law:
F = k * x
where,
F is the force applied,
k is the spring constant,
x is the displacement.
Rearranging the equation to solve for x:
x = F / k
Now, substituting the given values:
x = 0.0050 N / 1.20 N/m
Calculating the result:
x = 0.0042 m
Therefore, the tiny spring will be stretched to a displacement of 0.0042 meters (or 4.2 mm) when a 0.005 N force is applied.
To find the displacement of the tiny spring when a 0.005 N force is applied, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law equation: F = -kx
Where:
F = Force applied to the spring (in N)
k = Spring constant (in N/m)
x = Displacement of the spring (in m)
Now, we can rearrange the equation to solve for x:
x = -F / k
Plugging in the given values:
F = 0.005 N (force applied)
k = 1.20 N/m (spring constant)
x = -0.005 N / 1.20 N/m
Dividing -0.005 N by 1.20 N/m:
x ≈ -0.0042 m or -4.2 mm (approximated to four decimal places)
Therefore, the tiny spring will be stretched to a displacement of approximately -0.0042 m or -4.2 mm when a 0.005 N force is applied.