Recent content by splinter

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    Coin Rolling Up Incline, no mass given?

    Used KEi = PEf, solved for KEi and PEf, got KEi = .2186M, and PEf = 9.81MH. Since KEi = PEf, .2186M = 9.81MH, the mass on each side cancel out, so .2186 = 9.81H, then solve for H, i got .0223 meters, then divide that by sin(27) to get the value of the hypotenuse, which gave me .0491 meters for...
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    Coin Rolling Up Incline, no mass given?

    Coin Rolling Up Incline, no mass given!? 1. [SFHS99 8.P.46.] A coin with a diameter of 3.40 cm rolls up a 27.0° inclined plane. The coin starts with an initial angular speed of 55.0 rad/s and rolls in a straight line without slipping. How far does it roll up the inclined plane? Having some...
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    Force exerted to stop falling object

    Btw to find the time it took to slow to a stop, i used the kinematic equation Xf = Xi + 1/2(Vf+Vi)t
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    Force exerted to stop falling object

    ok, so I used -1/2mVf^2 to find the work done on the divers body, and got -5867.6 J. Then, since W = Fd, I divided the work done by the distance over which it was done, and got -1173.5 N. Tried both positive and negative of that, but online homework is still telling me its wrong!
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    Force exerted to stop falling object

    Having trouble with this problem: 3. [SFHS99 5.P.37.] A 50.0 kg diver steps off a 12.0 m high diving board and drops straight down into the water. If the diver comes to rest 5.0 m below the surface of the water, determine the average resistance force exerted on the diver by the water...
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    Solving "6. [SFHS99 5.P.60.]" Physics Problem

    Whoops it was work done by gravity...easy just mgd. doh! Thanks for the help i got them all now.
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    Solving "6. [SFHS99 5.P.60.]" Physics Problem

    Didn't see your new post and edit, might be able to get it with that.
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    Solving "6. [SFHS99 5.P.60.]" Physics Problem

    ok I've gotten the Fapp and the Fn, now I'm just having trouble with the Ff. I used the equation Ff = uN, plugging in my normal force of 194.5 and u of .3, then multiplied that by 3.1 to get work, and entered the negative of that, but it's still telling me its incorrect. Don't understand what...
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    Solving "6. [SFHS99 5.P.60.]" Physics Problem

    Ok so I tried taking the gravitational force, 54.96N, and finding the horizontal component of that. I drew it as a triangle with 54.96N as the y leg, then did 53.96/tan 30 to find the x leg of the triangle. This gave me 93.46N, which my online homework is saying is wrong! Still confused =\
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    Solving "6. [SFHS99 5.P.60.]" Physics Problem

    Hey everyone, my first post here, hope it's ok that it's a question... I've got this physics problem that is absolutely stumping me, is abnormal from all the other forces problems I've done before, and uk is throwing me for a loop too...here it is: 6. [SFHS99 5.P.60.] A 5.5 kg block is...
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