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AQA Mechanics 2 Question

Hi, can anyone help me with 11biii? Thanks
(edited 9 years ago)
Original post by Firefox23
Hi, can anyone help me with 11biii? Thanks


You need to apply F=ma.

Where F is the resultant force from the rope pulling up and the weight of the jumper pulling down.
Reply 2
Original post by ghostwalker
You need to apply F=ma.

Where F is the resultant force from the rope pulling up and the weight of the jumper pulling down.


How can there be a resultant force if the jumper is stationary, I tried to do that and flu that the force from the rope is equal to the weight
Original post by Firefox23
How can there be a resultant force if the jumper is stationary, I tried to do that and flu that the force from the rope is equal to the weight


The force on the rope is equal to the weight at the point of equilibrium, not at the point of maximum extension.

Jumper is only instantaneously stationary.
Reply 4
Original post by ghostwalker
The force on the rope is equal to the weight at the point of equilibrium, not at the point of maximum extension.

Jumper is only instantaneously stationary.


So will the force in the rope be equal to the elastic potential energy/distance?
Original post by Firefox23
So will the force in the rope be equal to the elastic potential energy/distance?


No - check the formulae for the EPE and for the force. It's not far off, if I interpret what you mean correctly.
Reply 6
Original post by ghostwalker
No - check the formulae for the EPE and for the force. It's not far off, if I interpret what you mean correctly.

Well Work done is force x distance and the EPE is the work done so to find the force just divide it by the distance travelled
Original post by Firefox23
Well Work done is force x distance and the EPE is the work done so to find the force just divide it by the distance travelled



"Work done = force x distance moved" tells you the work done by a constant force. If the force is varying, you'll need to use integration on the RHS, to find the work done.

Also, I'm not clear what force you're refering to here. You're posts are rather vague with the details.

From an energy consideration, the EPE will equal the loss in GPE in going from their starting position to their first stationary position (since the velocity will be zero in each position, the KE doesn't come into it).

The force in the rope is varying as the rope is stretched, so you can't used "work done = force x distance moved".

For EPE, you have work done in stretching the rope =λx22l= \dfrac{\lambda x^2}{2l} for any given extension.

You also have F=λxl F=\dfrac{\lambda x}{l} via Hooke's law, again for any given extension.

Those are the two formulae you should be comparing if you're going to do it that way.

Note: The extension of the rope is not the same as the distance fallen.
Reply 8
Original post by ghostwalker
"Work done = force x distance moved" tells you the work done by a constant force. If the force is varying, you'll need to use integration on the RHS, to find the work done.

Also, I'm not clear what force you're refering to here. You're posts are rather vague with the details.

From an energy consideration, the EPE will equal the loss in GPE in going from their starting position to their first stationary position (since the velocity will be zero in each position, the KE doesn't come into it).

The force in the rope is varying as the rope is stretched, so you can't used "work done = force x distance moved".

For EPE, you have work done in stretching the rope =λx22l= \dfrac{\lambda x^2}{2l} for any given extension.

You also have F=λxl F=\dfrac{\lambda x}{l} via Hooke's law, again for any given extension.

Those are the two formulae you should be comparing if you're going to do it that way.

Note: The extension of the rope is not the same as the distance fallen.


Oh yeah thanks, i used Hooke's law and that gave me the correct answer, i'll keep those equations and when to use them in mind, thanks so much

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