Question

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.

Answer 1

Answer:

Explanation:

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

Answer 2

In spring kinetic energy is directly proportional to the displacement. The spring energy of the given spring is 0.14375 J.  

The Kinetic Energy:

$U =\dfrac12 kx^2$

Where,

$U$ - potential energy

$k$ - Spring constant = 1.15 N/cm = 115 N/m

$x$ - displacement = 5 cm = 0.05 m

Put the values in the formula,

$U = \dfrac 12 115 \times 0.05 \\\\U = 0.14375\rm \ J$

 

Therefore, the spring energy of the given spring is 0.14375 J.

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