find the work done in extending a light elastic string to double its length

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Question Video: Elastic Strings and Springs

A ball of mass 1.8 kg is attached to one end of a light elastic string of natural length 2.4 m and modulus of elasticity 17.1 N. The other end of the string is fixed at a point 𝑂. The ball is released from rest at 𝑂. Taking 𝑔 = 9.8 m/s², find how far below 𝑂 the ball reaches before coming instantaneously to rest.

Question Video: Elastic Strings and Springs

Mathematics

A ball of mass 1.8 kg is attached to one end of a light elastic string of natural length 2.4 m and modulus of elasticity 17.1 N. The other end of the string is fixed at a point 𝑂. The ball is released from rest at 𝑂. Taking 𝑔 = 9.8 m/s², find how far below 𝑂 the ball reaches before coming instantaneously to rest.

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work done by stretching an elastic string

work done by stretching an elastic string

claref 3

OK, I have an issue with this...

the book says work done = [latex]\frac{1}{2}(T_1+T_2)(x_2-x_1)[/latex]

which is completely fine, it is the average tension (which is ok because tension doesnt increase quadratically or anything) multipled by the change in extension ie the distance travelled inthe direction of the stretching force.

but if the string starts off at its natural length, then the work done in stretching it to an extension of [latex]x_1[/latex] where the tension is [latex]T_1[/latex] is [latex]\frac{1}{2}T_1x_1[/latex], and to an extension of [latex]x_2[/latex] the work done would be [latex]\frac{1}{2}T_2x_2[/latex] (the book agrees up to here).

So, what i dont understand is why work done in stretching the string from an extension of [latex]x_1[/latex] to an extension of [latex]x_2[/latex] cannot be found by subtracting the work done in getting it to the extension of [latex]x_1[/latex] (from its natural length) from the work done in extending it (from nat. length) to [latex]x_2[/latex]. i think that should work because to get from the natural length to [latex]x_2[/latex] you have to strethch from the natural length to [latex]x_1[/latex] and then again from [latex]x_1[/latex] to [latex]x_2[/latex], so my way should work ie work done in stretching from an extension of [latex]x_1[/latex] to an extension of [latex]x_2[/latex] would be [latex]\frac{1}{2}(T_2x_2-T_1x_1)[/latex]

bu this is not the same as the first work done, why???

Reply 1 ttoby 15 claref

OK, I have an issue with this...

the book says work done = [latex]\frac{1}{2}(T_1+T_2)(x_2-x_1)[/latex]

which is completely fine, it is the average tension (which is ok because tension doesnt increase quadratically or anything) multipled by the change in extension ie the distance travelled inthe direction of the stretching force.

but if the string starts off at its natural length, then the work done in stretching it to an extension of [latex]x_1[/latex] where the tension is [latex]T_1[/latex] is [latex]\frac{1}{2}T_1x_1[/latex], and to an extension of [latex]x_2[/latex] the work done would be [latex]\frac{1}{2}T_2x_2[/latex] (the book agrees up to here).

So, what i dont understand is why work done in stretching the string from an extension of [latex]x_1[/latex] to an extension of [latex]x_2[/latex] cannot be found by subtracting the work done in getting it to the extension of [latex]x_1[/latex] (from its natural length) from the work done in extending it (from nat. length) to [latex]x_2[/latex]. i think that should work because to get from the natural length to [latex]x_2[/latex] you have to strethch from the natural length to [latex]x_1[/latex] and then again from [latex]x_1[/latex] to [latex]x_2[/latex], so my way should work ie work done in stretching from an extension of [latex]x_1[/latex] to an extension of [latex]x_2[/latex] would be [latex]\frac{1}{2}(T_2x_2-T_1x_1)[/latex]

bu this is not the same as the first work done, why???

Taking the first equation, we have work done = [latex]\frac{1}{2}(T_1+T_2)(x_2-x_1) = \frac{1}{2}(T_1x_2 - T_1x_1 + T_2x_2 - T_2x_1)[/latex] and we know that T=kx where k is the spring constant so

work done = [latex]\frac{1}{2}(kx_1x_2 - T_1x_1 + T_2x_2 - kx_2x_1)[/latex]

[latex]= \frac{1}{2}(T_2x_2 - T_1x_1 +kx_1x_2 - kx_2x_1)[/latex]

[latex]= \frac{1}{2}(T_2x_2 - T_1x_1)[/latex]

Reply 2 claref OP 3 ttoby

Taking the first equation, we have work done = [latex]\frac{1}{2}(T_1+T_2)(x_2-x_1) = \frac{1}{2}(T_1x_2 - T_1x_1 + T_2x_2 - T_2x_1)[/latex] and we know that T=kx where k is the spring constant so

work done = [latex]\frac{1}{2}(kx_1x_2 - T_1x_1 + T_2x_2 - kx_2x_1)[/latex]

[latex]= \frac{1}{2}(T_2x_2 - T_1x_1 +kx_1x_2 - kx_2x_1)[/latex]

[latex]= \frac{1}{2}(T_2x_2 - T_1x_1)[/latex]

ohh my god that is so clever thankyou sooo much!!

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Elastic Strings

Elastic strings A-Level Mechanics revision section looking at Elastic Strings.

Elastic Strings

Modulus and Natural Length

Elastic strings are strings which are not a fixed length (they can be stretched). Some strings are more stretchy than others and the modulus (or modulus of elasticity) of a string is a measure of how stretchy it is. The modulus is measured in newtons.

The length of an elastic string which does not have any forces acting upon it is known as the natural length of the string. If a string has been stretched, then the extension is how much longer the string is as a result of being stretched. Note that the extension = length of the string - natural length.

Hooke's Law

Hooke's law states that the tension in an elastic string (or spring), T, is found using the following formula:

, where l is the modulus of elasticity of the string, x is the extension of the string and l is the natural length of the string.

Example

A string with modulus (of elasticity) 10 N has a natural length of 2m. What is the tension in the string when its length is 5m?

T = 10 × 3 = 15 2

So the tension in the string is 15N.

Potential Energy Stored in String

When an elastic string is extended it has elastic potential energy.

Elastic potential energy stored in string = lx2/2l

In problems involving strings (and springs), if the only external force doing work is gravity then energy is conserved. Hence elastic potential energy + gravitational potential energy + kinetic energy = constant.

Springs

What has been said about strings also applies to springs. However, springs can be compressed as well as stretched. If a spring is compressed, then Hooke's Law still applies but T represents the "thrust" rather than the tension (basically the only difference is that thrust acts in the opposite direction to tension).

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