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Are There Any Physics Geeks?! i'm desperate!!

answers are fatalanswers are fatal Posts: 281
edited February 2009 in All Encompassing Trip
i have this physics problem that ive been trying to do forever and i can not comprehend it at all ! so if anyone can help that would be awesome!

the problem is ...

"Relative to the ground below, how many joules of PE does a 1000 N boulder have at the top of a 5 meter ledge? If it falls, with how much KE will it strike the ground?? What will be its speed on impact with the ground?"
thank you :)
Post edited by Unknown User on

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  • CreedDiseaseCreedDisease Posts: 241
    answers are fatal wrote:
    i have this physics problem that ive been trying to do forever and i can not comprehend it at all ! so if anyone can help that would be awesome!

    the problem is ...

    "Relative to the ground below, how many joules of PE does a 1000 N boulder have at the top of a 5 meter ledge? If it falls, with how much KE will it strike the ground?? What will be its speed on impact with the ground?"
    thank you :)
    Ok, Potential energy = mass * acceleration due to gravity * height (or PE=mgh)
    1000N is the mg part of the equation, the 5m is the h, so it is 5000Nm. 1Joule = 1Nm, so 5000 Joules.

    KE=1/2*m*v^2. to get m you first have to divide 1000N by 9.81 m/s^2. A N is 1kg*m/s^2, so you end up with mass (m) in kg. But you need v. v= acceleration * time. Distance x is acceleration * time^2/2.... So, solving for time, you get (2*x/a)^.5... so t=1.0096 sec.... v=9.81*1.0096 = 9.905 m/s

    KE=5000 Joules.
    Now that I did that, it occured to me that when the object hits the ground, this was to show the conservation of energy, PE=KE. From there you could have just solved for v since you know KE, and m. Either way, the math works and I don't feel like re-writing this... I hope this helps, and I hope I am right since i do this for a living.
  • CHANGEinWAVESCHANGEinWAVES Posts: 10,169
    CreedDisease wrote:
    answers are fatal wrote:
    i have this physics problem that ive been trying to do forever and i can not comprehend it at all ! so if anyone can help that would be awesome!

    the problem is ...

    "Relative to the ground below, how many joules of PE does a 1000 N boulder have at the top of a 5 meter ledge? If it falls, with how much KE will it strike the ground?? What will be its speed on impact with the ground?"
    thank you :)
    Ok, Potential energy = mass * acceleration due to gravity * height (or PE=mgh)
    1000N is the mg part of the equation, the 5m is the h, so it is 5000Nm. 1Joule = 1Nm, so 5000 Joules.

    KE=1/2*m*v^2. to get m you first have to divide 1000N by 9.81 m/s^2. A N is 1kg*m/s^2, so you end up with mass (m) in kg. But you need v. v= acceleration * time. Distance x is acceleration * time^2/2.... So, solving for time, you get (2*x/a)^.5... so t=1.0096 sec.... v=9.81*1.0096 = 9.905 m/s

    KE=5000 Joules.
    Now that I did that, it occured to me that when the object hits the ground, this was to show the conservation of energy, PE=KE. From there you could have just solved for v since you know KE, and m. Either way, the math works and I don't feel like re-writing this... I hope this helps, and I hope I am right since i do this for a living.
    ouch...my brain hurts after reading that. :|
    "I'm not present, I'm a drug that makes you dream"
  • redmosquito10redmosquito10 Posts: 568
    nicely put. any chance that physics knowledge translates to first-order system dynamics? :)
    "Ah, life is a gate, a way, a path to Paradise anyway, why not live for fun and joy and love or some sort of girl by a fireside, why not go to your desire and LAUGH..."
  • answers are fatalanswers are fatal Posts: 281
    CreedDisease wrote:
    answers are fatal wrote:
    i have this physics problem that ive been trying to do forever and i can not comprehend it at all ! so if anyone can help that would be awesome!

    the problem is ...

    "Relative to the ground below, how many joules of PE does a 1000 N boulder have at the top of a 5 meter ledge? If it falls, with how much KE will it strike the ground?? What will be its speed on impact with the ground?"
    thank you :)
    Ok, Potential energy = mass * acceleration due to gravity * height (or PE=mgh)
    1000N is the mg part of the equation, the 5m is the h, so it is 5000Nm. 1Joule = 1Nm, so 5000 Joules.

    KE=1/2*m*v^2. to get m you first have to divide 1000N by 9.81 m/s^2. A N is 1kg*m/s^2, so you end up with mass (m) in kg. But you need v. v= acceleration * time. Distance x is acceleration * time^2/2.... So, solving for time, you get (2*x/a)^.5... so t=1.0096 sec.... v=9.81*1.0096 = 9.905 m/s

    KE=5000 Joules.
    Now that I did that, it occured to me that when the object hits the ground, this was to show the conservation of energy, PE=KE. From there you could have just solved for v since you know KE, and m. Either way, the math works and I don't feel like re-writing this... I hope this helps, and I hope I am right since i do this for a living.



    thank u sooooooooooooooooooooooooooooooooooo much creeddisease!!! u made that sound so much easier than what my teacher "tries" to explain to the class.
  • soulsingingsoulsinging Posts: 13,202
    CreedDisease wrote:
    Ok, Potential energy = mass * acceleration due to gravity * height (or PE=mgh)
    1000N is the mg part of the equation, the 5m is the h, so it is 5000Nm. 1Joule = 1Nm, so 5000 Joules.

    KE=1/2*m*v^2. to get m you first have to divide 1000N by 9.81 m/s^2. A N is 1kg*m/s^2, so you end up with mass (m) in kg. But you need v. v= acceleration * time. Distance x is acceleration * time^2/2.... So, solving for time, you get (2*x/a)^.5... so t=1.0096 sec.... v=9.81*1.0096 = 9.905 m/s

    KE=5000 Joules.
    Now that I did that, it occured to me that when the object hits the ground, this was to show the conservation of energy, PE=KE. From there you could have just solved for v since you know KE, and m. Either way, the math works and I don't feel like re-writing this... I hope this helps, and I hope I am right since i do this for a living.

    I'm one semester from finishing law school, but I took AP physics in high school and almost became an engineer. I bailed to do English. Lately, I've been thinking I should have been an engineer... those kind of problems are kind of fun and refreshingly simple. There's a refreshing comfort in knowing that there's an answer... there's never a right answer in my field :(
  • redmosquito10redmosquito10 Posts: 568
    Don't doubt yourself too much, I'm in my third year of engineering school, and unfortunately things haven't been that clear-cut and simple for a while. Things get just as muddled, and most of the time there aren't answers unless you make certain assumptions or get data based on experimental evidence, etc. Nice, fun, simple physics ended after my 2nd physics class freshman year. oh well
    "Ah, life is a gate, a way, a path to Paradise anyway, why not live for fun and joy and love or some sort of girl by a fireside, why not go to your desire and LAUGH..."
  • Thoughts_ArriveThoughts_Arrive Melbourne, Australia Posts: 15,165
    answers are fatal wrote:
    i have this physics problem that ive been trying to do forever and i can not comprehend it at all ! so if anyone can help that would be awesome!

    the problem is ...

    "Relative to the ground below, how many joules of PE does a 1000 N boulder have at the top of a 5 meter ledge? If it falls, with how much KE will it strike the ground?? What will be its speed on impact with the ground?"
    thank you :)

    "Who's got the brain of J what's it mean to us now?"
    Adelaide 17/11/2009, Melbourne 20/11/2009, Sydney 22/11/2009, Melbourne (Big Day Out Festival) 24/01/2014
  • __ Posts: 6,651
    CreedDisease wrote:
    Ok, Potential energy = mass * acceleration due to gravity * height (or PE=mgh)
    1000N is the mg part of the equation, the 5m is the h, so it is 5000Nm. 1Joule = 1Nm, so 5000 Joules.

    KE=1/2*m*v^2. to get m you first have to divide 1000N by 9.81 m/s^2. A N is 1kg*m/s^2, so you end up with mass (m) in kg. But you need v. v= acceleration * time. Distance x is acceleration * time^2/2.... So, solving for time, you get (2*x/a)^.5... so t=1.0096 sec.... v=9.81*1.0096 = 9.905 m/s

    KE=5000 Joules.
    Now that I did that, it occured to me that when the object hits the ground, this was to show the conservation of energy, PE=KE. From there you could have just solved for v since you know KE, and m. Either way, the math works and I don't feel like re-writing this... I hope this helps, and I hope I am right since i do this for a living.

    Is it wrong for me to be a little turned on by this post? :o :oops: :ugeek:
  • CreedDiseaseCreedDisease Posts: 241
    Answers are fatal:
    Glad I could help :)

    Red Mosquito:
    Yes, I know dynamics. And things are even more muddled in the real world, but sounds like you have a good grasp on things. If you don’t have mathcad, you should try it out. It makes my professional life much easier.

    Soulsinging:
    I love engineering! I feel fortunate every day that I chose the path I took… Although I can always rely on math in engineering, there is not always a “right answer” situation. However, in Law, I assume there is always a right answer, you just have to convince everyone you have it.

    Scb:
    hahahaha, not wrong at all! Your post had the same affect on me ;)
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