1 A force og 10N causes a spring yo extend by 20mm.find A. the spring constant of the spring in N/m
B. The extension of the spring when 25N is applied C. The force applied that causes an extension of 5mm.
AAAaannndd the bot gets it wrong yet again!
A. To find the spring constant (k) in N/m, we can use Hooke's Law which states that the force applied on a spring is directly proportional to its extension. Mathematically, this can be expressed as:
F = k * x
where F is the force applied, k is the spring constant, and x is the extension of the spring.
In this case, we have a force of 10N and an extension of 20mm. We need to convert the extension to meters before calculating the spring constant:
x = 20mm = 20 / 1000 = 0.02m
Now, we can rearrange the equation to solve for the spring constant:
k = F / x
k = 10N / 0.02m
k = 500 N/m
Therefore, the spring constant of the spring is 500 N/m.
B. To find the extension of the spring when 25N is applied, we can rearrange the equation as follows:
F = k * x
We need to solve for x, so we rearrange the equation:
x = F / k
x = 25N / 500N/m
x = 0.05m
Therefore, the extension of the spring when 25N is applied is 0.05m (or 50mm).
C. To find the force applied that causes an extension of 5mm, we use the same equation:
x = F / k
Rearranging the equation to solve for F:
F = k * x
F = 500N/m * 0.005m
F = 2.5N
Therefore, the force applied that causes an extension of 5mm is 2.5N.
To find the spring constant (k) of the spring in N/m, we can use Hooke's Law, which states that the force applied to an elastic object (like a spring) is directly proportional to the extension or compression of the object.
Hooke's Law equation: F = k * x
Where:
F = Force applied to the spring
k = Spring constant
x = Extension or compression of the spring
A. To find the spring constant (k) in N/m:
Given:
Force (F) = 10N
Extension (x) = 20mm = 0.02m
Rearranging Hooke's Law equation:
k = F / x
Substituting the given values:
k = 10N / 0.02m
k = 500 N/m
Therefore, the spring constant of the spring is 500 N/m.
B. To find the extension of the spring when 25N is applied:
Given:
Force (F) = 25N
Rearranging Hooke's Law equation:
x = F / k
Substituting the given values:
x = 25N / 500 N/m
x = 0.05m = 50mm
Therefore, the extension of the spring when 25N is applied is 50mm.
C. To find the force applied that causes an extension of 5mm:
Given:
Extension (x) = 5mm = 0.005m
Rearranging Hooke's Law equation:
F = k * x
Substituting the given values:
F = 500 N/m * 0.005m
F = 2.5N
Therefore, the force applied that causes an extension of 5mm is 2.5N.
A. The spring constant of the spring in N/m = 0.5 N/m
B. The extension of the spring when 25N is applied = 50mm
C. The force applied that causes an extension of 5mm = 2.5N