A spring is extended 4mm by an applied force of 2.0N.
a)What is its spring constant?
b)How much would it extend if a load of 150g was suspended from it?
a. K = F/d = 2N/0.004m = 500 N/m.
b. F = m*g = 0.15kg * 9.8N/kg = 1.47 N.
(1.47N/500N) * 1m = 2.94*10^-3 m=2.94 mm
a) Well, the spring constant is like a superhero's strength - it tells us how much force it takes to stretch the spring by a certain amount. So, if a force of 2.0N stretches the spring by 4mm, we can calculate the spring constant by dividing the force by the displacement (extension). Using my amazing math skills, the spring constant would be 0.5 N/mm, or superhero strong!
b) Ah, the suspense is killing me! Okay, let's find out how much the spring will extend when a load of 150g is suspended from it. First, we need to convert 150g to Newtons because the spring loves the metric system. We all do, right? So, 150g is approximately 1.47N. Now we divide this load by the spring constant of 0.5 N/mm (the superhero spring constant we calculated before) and get a whopping extension of 2.94mm. So, the spring will spring into action and extend by 2.94mm.
a) To determine the spring constant, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension of the spring.
Hooke's Law can be expressed as:
F = -kx
Where:
F is the applied force
k is the spring constant
x is the extension of the spring
Given that the spring is extended by 4mm (or 0.004m) with an applied force of 2.0N, we can substitute these values into Hooke's Law to find the spring constant:
2.0N = -k * 0.004m
To solve for k, we divide both sides by 0.004m:
k = -2.0N / 0.004m
Therefore, the spring constant is:
k = -500 N/m (Note: The negative sign indicates that the spring force is opposite to the applied force.)
b) To determine how much the spring would extend if a 150g load (or 0.15kg) is suspended from it, we can use the formula:
F = -kx
Rearranging the equation to solve for x:
x = -F / k
Substituting the values into this equation:
x = -(0.15kg * 9.8 m/s^2) / (-500 N/m)
Simplifying:
x = 0.294m
Therefore, the spring would extend approximately 0.294m (or 29.4 cm) when a 150g load is suspended from it.
To find the spring constant (k) of the spring, you can use Hooke's Law, which states that the force applied to a spring (F) is directly proportional to the extension or compression of the spring (x).
Hooke's Law can be expressed as: F = k * x
a) To find the spring constant, we need to rearrange Hooke's Law equation:
k = F / x
Where:
- F is the applied force (2.0N)
- x is the extension of the spring (4mm)
Convert the extension from millimeters to meters:
- 4mm = 4 * 10^-3 meters
Substitute the values into the equation:
k = 2.0N / (4 * 10^-3 meters)
k ≈ 500 N/m
Therefore, the spring constant is approximately 500 N/m.
b) To find how much the spring would extend if a load of 150g (0.15kg) was suspended from it, we can use the same Hooke's Law equation:
F = k * x
Rearrange the equation to solve for x:
x = F / k
Where:
- F is the weight of the load (0.15kg multiplied by the acceleration due to gravity, which is 9.8 m/s²)
- k is the spring constant (500 N/m)
Substitute the values into the equation:
x = (0.15kg * 9.8m/s²) / 500 N/m
x ≈ 0.00294 meters
Therefore, the spring would extend approximately 0.00294 meters (or 2.94 mm) if a load of 150g was suspended from it.