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Work, Energy and Power Questions

  1. Oct 27, 2007 #1
    [SOLVED] Work, Energy and Power Questions

    Question 1
    A well trained soldier with a combat mass of 93 kg is able to jump from a height of 1.5 m and land safely by squatting down at a uniform rate as he lands. He squats down by 0.71 m. What force (in N) does his legs exert during the squatting process?

    Question 2
    There are 21 steps between each storey. The height of each storey is 2.8 m. From these data, estimate the efficiency of stair climbing (at 116 steps per minute) by the average person of 75 kg mass . (Hint: consider the gravitational potential energy gained and the energy consumed by a person climbing the stairs)

    Any help is appreciated:smile:
     
  2. jcsd
  3. Oct 27, 2007 #2
    The rules are - you have to show us your attempt to solve it.
    We will maybe give a hint. Get aware of Newton's Laws. They result in equations of motion. You need these to solve this.

    Break up the problem into parts. Figure the part where the soldier drops, up to just before he lands. Then figure the part after he contacts, and brings himself to a halt. Ask yourself what force it takes to do this. At all times, keep in mind all the forces on the soldier.
     
  4. Oct 27, 2007 #3
    For the 1st question i started off with finding the gravitational potential energy by using E = mgh but i got stuck.

    The 2nd question involves finding the total energy output divided by energy input to get the efficiency.

    Kinda stuck :S
     
  5. Oct 28, 2007 #4
    Still making no headway on this topic, anyone care to lend a helping hand?
     
  6. Oct 28, 2007 #5
    There are two ways you can get to the speed the soldier attained during the drop.

    One way is to know that all stuff, light or heavy, drops at the same rate. Comedy now - get fixed in your mind the image of Galileo dropping rocks from the Leaning Tower of Pisa, even if its an urban myth! The soldier's 93 kg is not going to make any difference to how fast he arrives! You don't need Newton for this bit. You only need to understand what an acceleration is.

    Something dropped 1.5 metres. The acceleration was g metres per second per second. The extra "per second" is not a typo. We say "metres per second squared". There are simple equations to figure how long it takes to drop. A time in seconds. Darn it though, you were given a distance instead, but thats OK, because wherever you find some equations of motion, you find a simple one that lets you plug in the drop time and get at the velocity of soldier splat.

    The "other" way is to say the potential energy m.g.h at the start will equal the kinetic energy (1/2)*m*v^2 when he arrives. You are after the 'v'. This is more direct, but I assure you it will not be enough to then figure the second part where the soldier is coming to a halt, involving forces and Mr. Newton.

    OK, so Google is your friend, and Wiki knows all. Now go to
    http://en.wikipedia.org/wiki/Equation_of_motion and look for "classic". Knowing that his initial velocity was zero helps, because the some equation terms become zero, leaving only an easy acceleration bit. Try and figure the velocity of arrival. Use that as the initial velocity and apply the equations again, but in reverse - this time to get at the acceleration that brought him to a halt in 0.71 metres.

    Getting the signs right is important here. If you arbitrarily choose downwards to be positive, then you will get a negative acceleration (a de-celeration?) that halts his motion. One of the numbers you need is his final velocity. (Yes - zero is a valid velocity!)
    To come to a halt, he gets accelerated upwards, even though his slowing motion was downwards. Go after that value of acceleration.

    Then, finally, invoke Newton. You know his mass. you figured the (de) acceleration. You can now figure the force on him.

    If you need to, come back with the equations of motion you tried. Show us what you did. You will find that knowing the concepts allows you to not bother learning the equations by heart. Finding the force on the soldier requires you understand the concepts. We will help you all we can to get you there. The final answer is just arithmetic, and we expect you can use a calculator for that, and it takes only a minute or two. This is an expansive answer. Try not to let us down.
     
  7. Oct 28, 2007 #6
    In fairness to you, it may be that you have learned about kinetic and potential energy, but not yet all about the equations of motion.

    You can solve this the fast way by simply staying with the kinetic_energy = potential_energy scheme all the way, without the need to go via equations of motion.

    Soldier gained kinetic energy in dropping 1.5m. This got dissipated in his legs bringing him to a halt in the shorter 0.71m. It is just as if his legs were springs while he chose to push. The amount of de-celeration he needed to halt in 0.71m compares to the acceleration g he had to speed up over distance 1.5m. They have in common the kinetic energy achieved at touchdown. Setting them equal allows you to discover you can simply scale g in proportion to the distances involved, to arrive at the (larger) acceleration involved in bringing soldier to a halt.

    Then you can bring in his 93kg, and Newton's law to figure the force involved.
     
  8. Oct 29, 2007 #7
    I understand the equating p.e to k.e part, but what about scaling the g in proportion? Would you mind elaborating?

    mgh = .5mv^2

    v^2 = 2X1.5g

    I assume the velocity value i get here is for the deceleration?
     
    Last edited: Oct 29, 2007
  9. Oct 29, 2007 #8
    I noticed that the answer could indeed be found via a calculating velocities, etc using the equations of motion. This may not be appropriate, and could possibly include work not yet expected of you, since there is a solution that can be understood using potential and kinetic energy considerations only.

    We happily discover that the whole business of calculating velocities can be ducked. In the end, you get to plug in easy numbers, but I shall expect you to show your working with enough feedback that you really have a grip on what is going on..
    OK - that starts off good. The first line sets potential energy (before) equal to kinetic energy (after).
    You notice the mass can cancel, and you have found the velocity (squared) that must be lost in the (de)celeration.
    At this point, things fizzle out. I thought you might attempt to figure the slowdown. This time, however, I deliberately lead you away from getting tangled in equations of motion.

    The soldier starts with potential energy
    [tex]PE=mgh_1 [/tex] where [tex] h_1=[/tex]1.5m

    He drops, gaining velocity [tex]v[/tex]. The potential energy has now all turned into kinetic energy..
    [tex]KE=\frac{mv^2}{2}[/tex]

    THEN something new happens. Just as he contacts the ground, he has all that kinetic energy.
    After a saggy 0.71m squat - its all gone!
    We don't much care where its gone. Even if all it did was heat up the soldier's muscles, the important thing was that it took a force enough to de-celerate in only 0.71m what had built up in a 1.5m drop with a acceleration of [tex]g[/tex]. We can call that (de)celeration [tex]a[/tex].

    Now, its not too tough to see why we can set the [tex]mgh_1=mah_2[/tex] where [tex]h_2[/tex] is the slowdown to halt distance 0.71m. The kinetic energy phase in between, and its associated velocity (squared) is just the stage in between. If the soldier had landed on a spring that compressed 0.71m, turning it all back into potential energy, the answer would have been the same.

    Thats as far as we go now. You should be able to see why I said you end up scaling 'g' in a certain proportion. Get that 'a'. Thats not enough! Use the a and Newton's law to answer the question. Please post your answers here. Also, let us know your thoughts about the reasoning.
     
  10. Oct 30, 2007 #9
    [tex]mgh_1=mah_2[/tex]
    93 x 9.81 x 1.5 = 0.71 x F
    F = 1927J

    Finally got it :D Thx for all the help
     
  11. Oct 27, 2010 #10
    Re: [SOLVED] Work, Energy and Power Questions

    the second question to simple .Just follow this...........
    energy stored as he reaches top is -MGH=21*2.8*75*9.8=43218
    now we need time...........
    for one step time is=60/116=0.517
    for 21 =0.517*21=10.86
    so power is =43218/10.86=3979.6
    or 3980 watt
     
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