IF A FORCE OF 100N STRETCHES A SPRING 0.1CM, FIND THE DONE IN STRETCHING THE SPRING 0.03CM . IF THE ELASTIC LIMIT IS NOTEXCEEDED?
F = k x
100 = k * 0.001 m
k = 10^5 N/meter
work = (1/2) k x^2 = (1/2) * 10^5 * (3 *10^-4 m)^2
= (1/2)(9) (10^-3) = 4.5 * 10^-3 Joules
To find the work done in stretching the spring by 0.03 cm, 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.
Hooke's Law can be written as:
F = kx
Where:
F is the force applied (in Newtons)
k is the spring constant (in Newtons per meter)
x is the extension or compression (in meters)
First, let's find the spring constant, k.
Given:
Force applied (F) = 100 N
Extension (x) = 0.1 cm = 0.1/100 = 0.001 m
Using Hooke's Law, we can rearrange the equation to find k:
k = F / x
k = 100 N / 0.001 m
k = 100,000 N/m
Now let's calculate the work done in stretching the spring by 0.03 cm.
Extension (x) = 0.03 cm = 0.03/100 = 0.0003 m
Work done (W) = (0.5)kx^2
W = (0.5)(100,000 N/m)(0.0003 m)^2
W = 0.5(100,000 N/m)(0.00000009 m^2)
W = 0.5(0.009 N)
W = 0.0045 N
Therefore, the work done in stretching the spring by 0.03 cm, without exceeding the elastic limit, is 0.0045 N.