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

Well, I think the following reasoning must be made.
\[\frac{1}{2}kx^2=\frac{1}{2}mv^2\Rightarrow v=x\sqrt{\frac{k}{m}}=3.68\cdot10^{-2}\sqrt{\frac{122}{0.172}}=0.98\ m/s\]

Answer 2

Where,
\[k=1.22\ N/cm=122\ N/m\]

Answer 3

I think the fact that the plane is inclined is not used here.

Answer 4

How did you get x?

Answer 5

oh wait never mind cm

Answer 6

But ya I have tried that answer before and it is incorrect not sure why

Answer 7

Then we need to put in some place the inclination. Do you have a picture of the problem?

Answer 8

I think there is a "strange" path to do it. Well, this is my proposal. This is an inclined plane, so there are two forces that acts in the block, the elastic force of the spring and the weight of the block. Applying second Newton's lawThis results in,
\[ma=kx-mg\sin\theta\Rightarrow a=\frac{kx-mg\sin\theta}{m}\]The block will travel a distance x=3.68 cm until the spring is completely free, so,
\[x=\frac{1}{2}at^2\Rightarrow t=\sqrt{\frac{2x}{a}}=\sqrt{\frac{2xm}{kx-mg\sin\theta}}\]Ans the velocity adquired by the block in this time is,
\[v=at=\sqrt{\frac{2x}{m}(kx-mg\sin\theta)}=1.31\ m/s\]

Answer 9

With all quantities in SI units,