What is the total work required to drive a screw into a block of wood?

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SUMMARY

The total work required to drive a screw into a block of wood is calculated using the concept of torque and its relationship with depth. The torque increases linearly from 0 N m to a maximum of 12 N m over 20 turns. The average torque is determined to be 6 N m, leading to a total work of 754 Joules when multiplied by the total distance turned. This solution effectively utilizes the linear relationship of torque and the definition of work in rotational motion.

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  • Understanding of torque and its calculation in rotational mechanics
  • Familiarity with the concept of work in physics
  • Basic knowledge of integrals and linear functions
  • Ability to perform calculations involving average values
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This discussion is beneficial for physics students, mechanical engineers, and anyone interested in understanding the principles of work and torque in mechanical systems.

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Homework Statement


It takes 20 turns to drive a screw completely into a block of wood. Because the
friction force between the wood and the screw is proportional to the contact area between
the wood and the screw, the torque required for turning the screw increases linearly with
the depth that the screw has penetrated into the wood. If the maximum torque is 12 N m
when the screw is completely in the wood, what is the total work (in Joules) required to
drive in the screw?


The Attempt at a Solution


So I tried to do an integral by finding out how much the torque changed per rotation. Then using that as the equation and the total distance turned to be plugged in.
\int.6x from 0 to 7200. I got a very large number and I don't think I did it right...
 
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Rather than use calculus, which is bound to throw you off, what is the average torque if it linearly increases? Then what is the definition of Work for a torque acting through one rotation?
 
Well, the equation for how it increases is just .6*(rotation number). The average torque would be the maximum and minimum divided by two...The maximum is obviously 12 so would that make the minimum be 0? That doesn't really make sense though. In that case the average would be 6. The equation for work is torque times the rotation distance. This gives 754 J which is the correct answer now that I check it. Thanks Phantom...guess I just needed another person to point me in the right direction!
 

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