Calculating Vertical Components in Ski-Jump Slopes

  • Thread starter Thread starter Claire84
  • Start date Start date
AI Thread Summary
The discussion revolves around solving mechanics problems related to an elastic string and ski-jump slopes. The first problem involves determining when the string becomes slack and how far the particle comes to rest, with emphasis on using the work-energy theorem to analyze motion and friction. The second problem addresses calculating vertical components for a jumper leaving a ski slope, where initial conditions and acceleration due to gravity are discussed. Participants suggest using energy conservation principles and clarifying the signs for acceleration to avoid complex numbers in calculations. The thread highlights the importance of correctly applying physics concepts to solve mechanics problems effectively.
Claire84
Messages
218
Reaction score
0
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 I've 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 I am 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
 
Physics news on Phys.org
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".
 
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.
 
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.
 
Time taken for the jumper to reach the ground after taking off horizontally from the ski-slope.
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top