The ball launcher in a pinball machine has a spring that has a force constant of 1.15 N/cm (Fig. P5.71). The surface on which the ball moves is inclined 10.0° with respect to the horizontal. If the spring is initially compressed 5.00 cm, find the launching speed of a 0.100 kg ball when the plunger is released. Friction and the mass of the plunger are negligible.

Given that,

Spring constant = 1.15N/cm

k=1.15N/cm

Let convert k to N/m

1m=100cm

So, k=1.15N/cm*100cm/1m=115N/m

Angle of inclination is 10°

Compression e = 5cm

e=0.05m

Mass of ball is m=0.1kg

Now, the energy in the spring is transferred to both kinetic energy and potential energy

Now, the Spring energy is given as

Us = ½kx²

Us = ½ * 115 * 0.05²

Us = 0.14375J

Kinetic energy is calculated as

K. E=½mv²

K. E=½*0.1*v²

K. E=0.05v²

Also, the potential energy is given as

P. E=mgh

The height with the transfer will be given using trigonometry

SinΘ = opp/hyp

SinΘ = h/e

h = eSinΘ

h = 0.05 Sin10

h = 0.00868m

Let, g=9.81m/s²

Then, mgh

P. E = 0.1*9.81*0.000868

P. E = 0.008517J

Now applying conservation of energy

Energy stored in spring = change in Kinetic energy plus change in potential energy

Us = ∆K. E + ∆P. E

Us=K. Ef - K. Ei + P. Ef - P. Ei

0.14375=0.05v² - 0 + 0.008517 - 0

0.14375 - 0.008517=0.05v²

0.135233=0.05v²

0.135233/0.05=v²

2.70466=v²

v=√2.70466

v=1.64m/s

So the launched speed is 1.64m/s