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Question about kinetic energy/toughness.

  • Thread starter mattyc33
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Another quantity recorded in the aforementioned research paper is the toughness of the spider silk,
found to be 27 J/cm^3 . Toughness is the amount of energy that an object can absorb, per unit volume of the object, before breaking.

Spiderman and Mary Jane freefall a height 66.9m before having their fall halted by a strand of web 1.00cm in diameter fired upwards a height 66.9m (so the strand of web is long). If all of their kinetic energy is absorbed by the spider silk, what is their maximum combined mass?

I think I understand the question. Our prof. has taught us about Tensile strength so I'm sure we're supposed to apply it here.

Tensile strength = F/A

What I did was change 27J/cm^3 to N/m^2 to get 2.7x10^7N/m^2.

Since they give you the length and width of the web, I converted 1.00cm to 0.01m and multiplied that by 66.9m, to get 0.669m^2.

They ask for the mass, and we've learned that in this case F = mg. Therefore I get an equation that looks something like this:

2.7x10^7N/m^2 = m(9.81)/0.669

Now I should be able to solve this easily but this gives me a huge mass which is physically impossible. I was wondering if anyone would be able to show me my error. Thanks alot.
 

Answers and Replies

  • #2
PeterO
Homework Helper
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Another quantity recorded in the aforementioned research paper is the toughness of the spider silk,
found to be 27 J/cm^3 . Toughness is the amount of energy that an object can absorb, per unit volume of the object, before breaking.

Spiderman and Mary Jane freefall a height 66.9m before having their fall halted by a strand of web 1.00cm in diameter fired upwards a height 66.9m (so the strand of web is long). If all of their kinetic energy is absorbed by the spider silk, what is their maximum combined mass?

I think I understand the question. Our prof. has taught us about Tensile strength so I'm sure we're supposed to apply it here.

Tensile strength = F/A

What I did was change 27J/cm^3 to N/m^2 to get 2.7x10^7N/m^2.

Since they give you the length and width of the web, I converted 1.00cm to 0.01m and multiplied that by 66.9m, to get 0.669m^2.

They ask for the mass, and we've learned that in this case F = mg. Therefore I get an equation that looks something like this:

2.7x10^7N/m^2 = m(9.81)/0.669

Now I should be able to solve this easily but this gives me a huge mass which is physically impossible. I was wondering if anyone would be able to show me my error. Thanks alot.
It is not enough to merely match gravity [mg]. That just means the object stops accelerating. Since they have already fallen 66+ metres, they will be travelling at about 35 m/s. It is not enough to stop accelerating, they need to be brought to a halt!!

If the upward force from the strand was 2mg [an excess upwards force of mg] they will take a further 66+ metres to stop. [a net downward force of mg took 66+ metres for them to gain their speed, a net upward force of mg will take just as long to stop them again.

How quickly did you want to stop them?
 

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