Recent content by strategist

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    Undergrad Why does bringing an object closer to Earth do negative work

    Ah okay. So if they ask "how much work did you do to stop the object", what kind of information would you think I'd be provided? I just don't want to get caught off guard by something like this and wind up missing it on the test. The mass and radius alone wouldn't be sufficient, would they?
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    Undergrad Why does bringing an object closer to Earth do negative work

    I think that is what they're implying actually. It's a little weird to think about. Would it be like I'm consistently pushing upwards on a falling object but that force I'm applying is less than the force of gravity? Either way, I hope something like this isn't on the SAT Physics subject test. I...
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    Undergrad Why does bringing an object closer to Earth do negative work

    Sorry, but I don't really understand what you mean by the "no work is done against the field". I mean I understand that letting the object just fall requires no work but why does just "allowing" the object to fall mean we're doing negative work? In my mind, that sort of sounds like we're doing...
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    Undergrad Why does bringing an object closer to Earth do negative work

    I'm reading through Sparknotes' explanation of the equation: U = -Gm1m2/r "Off the surface of the Earth, there’s no obvious reference point from which to measure gravitational potential energy. Conventionally, we say that an object that is an infinite distance away from the Earth has zero...
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    Undergrad Is mass relevant on a frictionless surface?

    Oh, I understand that. But what I'm asking is if the mass were greater and the same force were applied, would the velocity be any different?
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    Undergrad Is mass relevant on a frictionless surface?

    I was looking at a problem: http://gyazo.com/c872ea999197823a42568809f9d97d3f and I understood that the reason that the force would have to be greater on a surface with friction because the equation for the force of friction is dependent on mass (μk * mg) and with two masses it essentially...