Some quick h/work help please!

  • Thread starter Claire84
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  • #1
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Hi, I'm just doing my Applied Maths h/worl at the mo which is pretty much just mechnics and I was hoping someone here could give me a hand. I think my problem here is more a case of my brain shutting down early for rimbo as opposed to the h/work being difficult so here I go.......

The first problem I'm having is this one-

One end of an elestic string of natura; length l and modulus lambda is fixed to a point O on a rough horizontal table (its 'spring constant' is k/lambda). The other end is attached to a particle P of mass m. The system is released from rest with P lying on the table and the distance OP being 2l. If lambda=3mg and u (coefficient of friction)=1 between the table and the particle, show that-
a) the dtring eventually becomes slack
b)the particle comes to rest at a distance l/2 from O.

For part a Ive been using the work energy theorum because if at B (a distance l from O, as in at the natural length), the velocity is greater than 0 then the particle will keep on moving so the string will become slack. so I'm using KE+PE at P minus KE+PE at B= work done my the frictional force. So at P we'd only have PE since it's at rest and is it true that at B we'd only have KE since the particle would be at the string's natural lenth therefore the string wouldn't be strectched in anyway? I've got work done by the frictional force as -mgl which I'm assuming is ok. My problem is hat Im end up with my velocity at B to be the square root of 5gl, which I have a feeling is wrong because for part b I'm not getting the ansewet to be l/2. For part b I'm using rhe fact that there's no tension so only friction acts to slow the particle down so I'm using the eqt of motion v^2=u^2+2as, but I'm getting 5l/2 out of this. Can anyone shed any lioght on this for me?

The next one is just short. If we have a ski-jump slope and the jumper goes off the bottom of it moving horitzonally, how do you work out the verticla components? I've got u=0, a=g and s=-h (of h is the hgith above the ground when taking off). However if you put that into an ewqt of motion you end up with a complex number so what is going wrong?

Claire
 

Answers and Replies

  • #2
Doc Al
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Originally posted by Claire84
For part b I'm using rhe fact that there's no tension so only friction acts to slow the particle down so I'm using the eqt of motion v^2=u^2+2as, but I'm getting 5l/2 out of this. Can anyone shed any lioght on this for me?
Use the "work-energy" theorem for part b as well. Initial energy is spring potential energy (KE = 0); at final position, energy = 0.
If we have a ski-jump slope and the jumper goes off the bottom of it moving horitzonally, how do you work out the verticla components? I've got u=0, a=g and s=-h (of h is the hgith above the ground when taking off). However if you put that into an ewqt of motion you end up with a complex number so what is going wrong?
What are you trying to find? The speed of the jumper after going down a hill of height h? (If so, assume no friction: energy is conserved.) Not sure what you mean by "vertical components".
 
  • #3
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If I was trying to find time by resolving vertically at he end of the ski slope. Like if I were resolving horizontaly there'd be no acceleration etc.
 
  • #4
Doc Al
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Originally posted by Claire84
If I was trying to find time by resolving vertically at he end of the ski slope. Like if I were resolving horizontaly there'd be no acceleration etc.
Time of what? I still don't know what the problem is.
 
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Time taken for the jumper to reach the ground after taking off horizontally from the ski-slope.
 
  • #6
Doc Al
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Now I see what you are doing.

Originally posted by Claire84

I've got u=0, a=g and s=-h (of h is the hgith above the ground when taking off). However if you put that into an ewqt of motion you end up with a complex number so what is going wrong?
If down is negative, then the acceleration is -g. That should fix your problem.
 

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