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Surviving the MCAT

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HELP POOR PHYSICS STUDENT [Mar. 13th, 2008|07:50 pm]
Surviving the MCAT
mcat_survival
[skinnyhunny2]
A human canon has a spring constant of 35000N/m. The spring can be extended up to 4.5 m. How far (horizontally) would a 65kg clown be fired if the cannon is pointed upward at 45 degrees to the horizontal??

I CANT DO IT !!!!!!! HELP!!!!!!!!!! the answer is 1.1X10^3 m
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Comments:
[User Picture]From: ucla_bruin
2008-03-24 08:01 pm (UTC)
you prolly already have the method devised for this, but you can find the force exerted by the spring using F=kx since the spring is static. Plug in the spring constant and the displacement and solve for F. Using the force, you have the mass of the clow (65kg * 9.8) and you can find acceleration. Using the acceleration you can probably solve using kinematics. The horizontal component (some velocity)cos45 plugged into either X=Xo+(Vo)t+1/2(a)(t^2) or Vf^2=Vi^2+2a(Xf-Xi) (depending on how you solved) should give you the answer you're looking for.
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From: (Anonymous)
2008-12-23 05:49 pm (UTC)

your physics question

hey buddy, I think ive got your answer

Use conservation of energy to find the initial velocity
(1/2kx^2+1/2mv^2)initial = (1/2kx^2+1/2mv^2)final
solve for vfinal (two of the terms in the equation cancel out)
Once you have that, break the velocity vector into components and use the y component V0ysin(45) to find the height the clown will go using the kinematic equation Vf=Vi+at in this case a will be -g free fall due to gravity. Once you know t, multiply it by two since it will take twice the amount of time at max height to come back down to earth and use the only relevant equation in the x direction to find the range which is x=vt and you get 1107 meters.
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