A spring 20cm long is stretched to 25cm by a load of 50newton.what will be it's length by 100newton.
Assuming that the spring follows Hooke's law (that the force applied is directly proportional to the extension of the spring), we can set up a proportion to solve for the new length:
(load 1/load 2) = (extension 1/extension 2)
Where:
load 1 = 50 N
extension 1 = 25 cm - 20 cm = 5 cm
load 2 = 100 N
extension 2 = ?
Plugging in the values:
(50 N/100 N) = (5 cm/extension 2)
Simplifying and solving for extension 2:
(1/2) = (5 cm/extension 2)
Multiplying both sides by extension 2:
extension 2/2 = 5 cm
extension 2 = 10 cm
Therefore, the length of the spring by 100 N will be 20 cm + 10 cm = 30 cm.
To answer this question, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring.
First, let's calculate the spring constant (k) using the formula:
k = F / x
where:
k = spring constant
F = applied force
x = extension or compression of the spring
Given:
F1 = 50 Newtons
x1 = 25 cm - 20 cm (extension)
F2 = 100 Newtons (new force)
x2 = unknown (new extension)
Calculating the spring constant (k):
k = F1 / x1 = 50 N / (25 cm - 20 cm) = 50 N / 5 cm = 10 N/cm
Now, let's calculate the new extension of the spring (x2) using the spring constant (k) and the new force (F2):
x2 = F2 / k = 100 N / 10 N/cm
Converting cm to meters (1 cm = 0.01 m):
x2 = (100 N / 10 N/cm) × (1 cm / 0.01 m) = 1000 cm/m
Therefore, the length of the spring by 100 Newtons will be 1000 cm or 10 meters.