A force of 10N causes aspring to 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
Recall that F = kx. So,
(a) 10 = k * 0.02
(b) 25 = kx
(c) F = k * 0.005
Student
To find the spring constant, we can use Hooke's Law equation, which states that the force applied to a spring is directly proportional to the extension or compression of the spring.
a) The equation for Hooke's Law is F = kx, where F is the force applied to the spring, k is the spring constant, and x is the extension or compression of the spring.
Given that a force of 10N causes a spring to extend by 20mm, we can substitute these values into Hooke's Law equation:
10N = k * 20mm
First, we need to convert the extension from millimeters (mm) to meters (m) to ensure that the units are consistent:
20mm = 20/1000 m = 0.02 m
Now we can rewrite the equation:
10N = k * 0.02m
To solve for k (the spring constant), divide both sides of the equation by 0.02m:
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 use the same equation, F = kx, and rearrange it to solve for x:
25N = 500 N/m * x
To find x, divide both sides of the equation by 500 N/m:
x = 25N / 500 N/m
x = 0.05 m
Therefore, the extension of the spring when 25N is applied is 0.05 meters (or 50 mm).
c) To find the force applied that causes an extension of 5mm, we can rearrange the Hooke's Law equation to solve for F:
F = kx
Given x = 5mm, we need to convert it to meters as before:
5mm = 5/1000 m = 0.005 m
Now we can substitute the values into the equation:
F = 500 N/m * 0.005 m
F = 2.5 N
Therefore, the force applied that causes an extension of 5mm is 2.5 N.