Such as in the case of loading with a load w which after removing the spring comes back to its original shape and size this deformation in which the body regains its shape and size after removing the load is called elastic deformation when the body is loaded with a load w there will be a resisting force that develops inside a body when we remove the load this resisting force makes the body regains its original shape and size on the other hand plastic deformation is the deformation of a body that permanently changes the shape of the body when an applied force is very high and a body cannot handle it in this case the body goes under permanent deformation generally all the objects show both elastic and plastic deformations for example when we load a spring with a certain load first it will show elastic deformation up to some limit but if we go on adding more load the spring will break and deform permanently in this case the elastic limit of a wire depends on the nature of a material and the diameter of the spring as we understood what is elastic and plastic deformations in 1660 an English physicist called Robert hook stated this law which states for elastic deformation in spring the deforming force is directly proportional to the change in length consider this spring whose one end is fixed and has a length l the other end is loaded with load w we can see that the spring undergoes deformation and its length increases let's call this increase in length delta the hook's law states that the load w hanging on the spring or the deforming force applied on the spring due to load w is directly proportional to the change in length of the spring that means if the deforming force changes then the length of the elongation of the spring also changes proportionally in other words the deforming force is equal to some constant times change in length of the spring therefore we can write the deforming force f is equal to k times delta l this is how mathematically Hooke's law is denoted one important thing to note here is this Hooke's law is only valid within the elastic limit of the body where the body regains its original form after removing the deforming force this law is not only valid for springs but is valid for elastic deformation in any material 'but' there is an assumption that states this law is only valid when the forces and deformations are relatively small the modified version of Hooke's law states stress is directly proportional to strain within elastic limits that means when stress increases the strain also increases with some proportions if we remove the stress the spring regains its original size this behavior is up to a certain point which is within this elastic region here the constant of proportionality is e where e is called young's modulus or simply modulus of elasticity this equation can be rearranged as e equal to stress divided by strain the value of e is different for different materials which is calculated as the ratio of stress applied on a body to the strain undergone by the body the unit of young's modulus is the same as stress that is newton per mm square or newton per meter squared depending on the unit used for calculating area a weight of 12 newtons is hung on a helical spring which has an unloaded length of 16 centimeters now a helical spring is just your standard spring which is looking like it's coiled like that that's a helical like a helix so an unloaded length of 16 centimeters and calculate the value of the spring's spring constant in this situation here we're loading it with 12 newtons 'and' it's got a new length of 21 centimeters so let's get stuck into this calculation now firsts we need to find the extension of the spring and that's fairly straightforward because it's just the difference between the original length and the new lengths we would do 21 minus 16 and that would give us an extension x equals 5 centimeters that's quite an important point sometimes people use the new length of the spring in the calculations 'and' it all goes wrong you've got to use the extension so let's make a data list to start with what do we know well we know that the force applied is 12 newtons we know that the extension of the spring is 5 centimeters 'and' we're being asked to find the spring constant k which is the stiffness of the spring so that's what we're looking for and that will have the units of newton per centimeter because we're using extension in centimeters then we will get the spring constant in newton per centimeter they have to agree okay the equation is f equals k times x firsts we've listed our data we've then written down the equation which links all the data together 'and' now we're going to solve we're going to put the numbers in let's do that we've got 12 equals k multiplied by 5. I'm, using a bracket here for multiplication we need to get k on its own we could divide both sides by 5 we've got 12 divided by 5. If we put that into the calculator we'll find that that comes to 2.4 is equal to k, and finally we can state the answer with the correct unit of our spring constant k
is 2.4 and check the units up here newton per centimeter and that's how you would do a calculation using Hooke's law f equals k x ‚ and just make sure that you use the extension not the lengths of the spring it's the extension which is the important
Quantity for x a weight of 12 newtons is hung on a helical spring which has an unloaded length of 16 centimeters now a helical spring is just your standard spring which is looking like it's coiled like that that's a helical like a helix so an unloaded length of 16 centimeters and calculate the value of the spring's spring constant in this situation here we're loading it with 12 newtons 'and' it's got a new length of 21 centimeters so let's get stuck into this calculation now firsts we need to find the extension of the spring and that's fairly straightforward because it's just the difference between the original length 'and' the new length 'so' we would do 21 minus 16 and that would give us an extension x equals 5 centimeters that's quite an important point sometimes people use the new length of the spring in the calculations 'and' it all goes wrong you've got to use the extension so let's make a data list to start with what do we know well we know that the force applied is 12 newtons we know that the extension of the spring is 5 centimeters 'and' we're being asked to find the spring constant k which is the stiffness of the spring 'so' that's what we're looking for and that will have the units of newton per centimeter because we're using extension in centimeters then we will get the spring constant in newton per centimeter they have to agree okay the equation is f equals k times x so firsts we've listed our data we've then written down the equation which links all the data together 'and' now we're going to solve we're going to put the numbers in let's do that 'so' we've got 12 equals k multiplied by 5. Using a bracket here for multiplication, 'so' we need to get k on its own 'so' we could divide both sides by 5. 'So' we've got 12 divided by 5. If we put that into the calculator we'll find that that comes to 2.4 is equal to k and 'so' finally we can state the answer with the correct unit our spring constant k
is 2.4 and check the units up here newton per centimeter and that's how you would do a calculation using Hooke's law f equals k x, and just make sure that you use the extension not the lengths of the spring it's the extension which is the important
quantity for x
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