final velocity of the pin ball.

Question

K=60 N/m

pull back the pin 5cm

friction: 0.2

and mass is 0.08kg.

θ: 12

side 1 is 44.13cm. the side opposite of that is 49.13cm. the length is 88.25 cm

(a.k.a height change is 5cm)

Determine the speed of the pinball when it reaches the top of the table after you shoot it. Note: To make life simpler, assume when the spring is compressed, that is flat against the front of the machine.

if I am missing information, please let me know and i will update it. Thank you

Answer 1

Using the information given, the pinball actually does not have enough energy to reach the top of the table.

Given:

k = 60 N/m

Spring compression = 5 cm = 0.05 m

m = 0.08 kg

Coefficient of friction μ = 0.20

θ = 12°

Distance along table = 88.25 cm = 0.8825 m

First find the energy stored in the spring:

Spring energy = (1/2)kx²

= (1/2)(60)(0.05)²

= 0.075 J

Now calculate the work done against friction:

Friction force = μmg cos(θ)

= (0.20)(0.08)(9.81)cos(12°)

≈ 0.154 N

Work against friction:

W = Fd

= (0.154)(0.8825)

≈ 0.136 J

Notice that the spring starts with only 0.075 J of energy, while friction alone would require about 0.136 J to travel the entire distance.

0.075 J < 0.136 J

Therefore, even before accounting for the increase in gravitational potential energy as the ball moves uphill, the spring does not provide enough energy for the pinball to reach the top.

So, with the numbers provided:

The pinball will stop before reaching the top, so there is no final velocity at the top.

One thing to double-check is the given geometry. A table length of 88.25 cm at an angle of 12° gives a height change of about 18.3 cm, not 5 cm. However, even if the height change really is only 5 cm, the pinball still does not have enough spring energy to overcome the stated friction.