A spring 20cm long is stretched to 25cm by a load of 50N. What will be it's length when stretched by 100N assuming that the elastic limit is not reached?
We can use Hooke's law which states that the extension of a spring is directly proportional to the force applied, as long as the elastic limit is not reached.
Let x be the extension of the spring when a load of 50N is applied. Using Hooke's law, we have:
x/20 = 50/ƙ where ƙ is the spring constant
Solving for ƙ, we get:
ƙ = 50/(x/20) = 1000/x
Now we can use this value of ƙ to determine the extension of the spring when a load of 100N is applied:
x/20 = 100/ƙ
Substituting ƙ = 1000/x, we get:
x/20 = 100x/1000
Simplifying, we find:
x = 25cm
Therefore, the spring will be stretched to 25cm + 5cm = 30cm when a load of 100N is applied.
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, as long as the elastic limit is not reached.
Hooke's Law equation: F = k * x
Where:
F is the force applied to the spring
k is the spring constant (a measure of how stiff the spring is)
x is the extension or compression of the spring
We can rearrange the equation to solve for the extension or compression:
x = F / k
To find the spring constant, we can use the given information that the spring is stretched from 20cm to 25cm by a load of 50N:
x = 25cm - 20cm = 5cm = 0.05m
F = 50N
Using Hooke's Law, we can calculate the spring constant:
k = F / x
k = 50N / 0.05m
k = 1000 N/m
Now, we can use the spring constant to find the new extension when the load is 100N:
F = 100N
x = F / k
x = 100N / 1000 N/m
x = 0.1m
Converting the extension to centimeters gives us:
x = 0.1m * 100cm/m
x = 10cm
Therefore, the length of the spring when stretched by 100N will be 10cm.