A spring of natural length 30.0cm is stretched 35.0cm by a load of 15N Cal calculate the extension produce
Pls answer nowwww
As written the question makes no sense. Please proofread it.
impatient much?
Hard to say.
Did it stretch from 30 to 35 cm -- stretched by 5 cm ?
or did it stretch by 35cm, from 30 to 65 cm ?
Or is this just a trick question? What does the force of 15N have to do with it?
Note: F = kx
Maybe you can use that
e =l1-l0=35-30=5cm
Based on the given information, the extension produced is 5 cm.
To calculate the extension produced by the load on the spring, you need to use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it.
Hooke's Law is represented by the equation:
F = kx
Where:
F is the applied force
k is the spring constant
x is the extension of the spring
In this case, the applied force (F) is 15N, and the natural length of the spring (L) is 30.0cm, which means the spring is stretched by 35.0cm - 30.0cm = 5.0cm.
The spring constant (k) is not given in the question, which means we may need additional information to calculate it. The spring constant is specific to each spring and measures its stiffness. It represents the force required to stretch or compress a spring by a certain distance.
If the spring constant is provided, you can substitute it into the equation F = kx, rearrange the equation to solve for x, and find the extension produced.
If the spring constant is not provided, you won't be able to calculate the extension with the given information.