Question

The ball launcher in a pinball machine has a
spring that has a force constant of 1.22 N/cm.
The surface on which the ball moves is in-
clined 15.8 with respect to the horizontal.

If the spring is initially compressed 3.68 cm,
find the launching speed of a 0.172 kg ball
when the plunger is released. The acceleration
due to gravity is 9.8 m/s2 . Friction and the
mass of the plunger are negligible.
Answer in units of m/s

Answer 1

Work done = integral of spring force minus component of gravity down incline \[\Large \int\limits(-kx -0.172*9.8 \sin 15.8)dx\]

hmmm after writing that, maybe you don't need an integral. Just find the net work done (work is force times distance:

Work done by spring = 1/2*kx^2 where x=0.0368 m (metres not cm)
work done by gravity = 0.172*9.8*sin15.8*x where x=0.0368m

convert k=1.22 N/cm to 122N/m and calculate net work (spring work minus gravity)

Net work done = 1/2*122(0.0368)^2 - 0.172*9.8*sin15.8*0.0368
=0.08831 Joules

Answer 2

work done = change in KE = 1/2*mv^2
where m = 0.172

gives v = 1.01 m/s

Answer 3

I understand how you got that answer but it is incorrect for this problem

Answer 4

Hmm, idk then :(

Answer 5

It's ok :) it's a sucky problem lol

Answer 6

I'll try to figure it out, maybe i have a number wrong... i'm pretty sure the method is right.

Answer 7

Sounds good I've tried different wAys to get it but it hasn't worked

Answer 8

Do you know what the answer is?

Answer 9

this problem is solved EXACTLY the same way i solved your problem: http://www.csupomona.edu/~skboddeker/131/131hw/ch7h.htm

scroll down to:

Ch 7 #63

Answer 10

I don't know what the answer is :/
But ya that's what I'm saying I've tried

Answer 11

So at this point i wonder if the "correct" answer to the problem is wrong... if you and i have tried multiple times, and my method is exactly the same as the one above... I'm checking all my numbers just to be sure but it looks all correct.

Answer 12

Ask your lecturer or whoever...

Answer 13

Ha ha imma have to thanks for your help though :)

Answer 14

No prob. Let me know what you find out... cos i can't find any errors, and i've looked up a few similar questions, all solved the same.

Answer 15

@RavenLynette i found the error haha... i used wolfram alpha to find the work done, and it used radians not degrees!

Answer 16

1/2*122(0.0368)^2 - 0.172(9.8)(0.0368*sin(15.8))

Net work done = 1/2*122(0.0368)^2 - 0.172*9.8*sin15.8*0.0368
=0.0657191 Joules (using degrees this time not radians...)

work done = change in KE = 1/2*mv^2 = 0.0657191

v = 0.874 m/s

Answer 17

Haha omg thank you :3

Answer 18

you're welcome! :D