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martindcx1e
Old 03-05-2008, 02:40 AM     Post subject: Another physics problem #1 (permalink)  
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Hey all I'm havin some troubles with physics again. Maybe someone here knows how to do this...

1) The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm. The surface on which the ball moves is inclined 10.0° with respect to the horizontal. If the spring is initially compressed 4.00 cm, find the launching speed of a 70 g ball after the plunger is released and fully extended. Friction and the mass of the plunger are negligible.



Ok so I need to change things to be in terms of m and kg. So 1.20N/cm = 120N/m, 4cm = .04m, and 70g = .07kg.
k = 120
x = .04
m = .07
g = 9.8

Now I use the Work-Kinetic Energy Theorem which is W_net = DeltaK
DeltaK is final K - initial K and K = .5mv^2. The initial K is 0 since v = 0 there so all I need is final K which is (.5)(.07)(v^2)

Now the other side of the equation is W_net which is W_spring + W_gravity. W_spring = -(.5)(k)(x^2) where x is how much the spring is compressed. This = -(.5)(120)(.04^2) which is -.096. W_gravity = (m)(g)(height) which is (.07)(9.8)(sin10) = .119. So W_s + W_g = .119-.096 = .023

Set .023 equal to final K which is (.5)(.07)(v^2) which says .023 = .035v^2 and then divide .023/.035 = .657 and then take sqrt (.657) = .81 which is v.

WTF am I doing wrong?!?
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Pelion
Old 03-05-2008, 08:49 AM #2 (permalink)  
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Start by working up how much energy you have stored up in the spring.

Then work out how much of that is going to be converted into gravitational potention energy when you move 4cm and 10deg.

Whatever is left over will go into the kinetic energy of the ball.
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