# Homework Help: Spring Questions about ball launcher

1. Dec 4, 2006

### dustpal

I need help ASAP cuz the teacher assigned it today and its due tomorrow....

1. The problem statement, all variables and given/known data
The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm. The surface on which the ball moves is inclined 10.0° with respect to the horizontal. If the spring is initially compressed 4.50 cm, find the launching speed of a 80 g ball when the plunger is released. Friction and the mass of the plunger are negligible.

2. Relevant equations
Fs = -kx
Ws = change in KE
KE = 0.5mv^2
w = mg

3. The attempt at a solution

w(x)=mg*sin(10) >>>>>> w(x)=0.136 N
Fs = -(1.2)(-4.50) >>>>> Fs = 5.4 N
F(netx)= Fs - w(x) >>>> 5.4 - 0.136 >>>>> F(netx) = 5.264 N
Ws = Fd >>>> Ws = 5.264*0.0450 >>>>> Ws = 0.23688 J
Ws = 0.5mv^2 >>>>> 0.23688 = 0.5(0.08)V^2 >>>>> v = 2.43 m/s (which is about twice as much as it should be)

1. The problem statement, all variables and given/known data
A light spring with spring constant k1 is hung from an elevated support. From its lower end a second light spring is hung which has spring constant k2. An object of mass m is hung at rest from the lower end of the second spring.
(a) Find the total extension distance of the pair of springs. (Use k1 for k1, k2 for k2, and m and g as appropriate in your equation.) (b) Find the effective spring constant of the pair of springs as a system. We describe these springs as in series. (Use k1 for k1 and k2 for k2 as appropriate in your equation.)

2. Relevant equations
Fs = -kx
w = mg
Ws =Fd = change in KE
KE = 0.5mv^2

3. The attempt at a solution

Honestly, we didn't really cover springs all too well in our physics class so i have no clue where to even begin, just a hint would help me so much. The one example we did was on a horizontal plane so that's why i think i'm having difficulties doing vertical and inclined springs.

2. Dec 4, 2006

### PhanthomJay

In the first problem, it is best to use energy methods, but if you do it your way, you must first note that the 5.4N spring force is the maximum force exerted by the spring. Its average force is 1/2 that amount. Secondly, your formula for Ws =Fd = 1/2 mv^2 is an incorrect application of the work energy theorem, not only because you must use the average force, but also because you neglected the work done by gravity. So rather than get bogged down with average forces and work, use the conservation of energy theorem
$$PE_{spring}_{initial} + PE_{ball}_{initial} + KE_{ball}_{initial} = PE_{spring}_{final} + PE_{ball}_{final} + KE_{ball}_{final}$$ . Do you know how to calculate or find each of these terms?

Last edited: Dec 4, 2006
3. Dec 5, 2006

### dustpal

well....we aren't allowed to use energy conservation yet because we haven't learned it. If u could help correct what i have, that would be the best way. But i don't understand what u mean by avaerage force and stuff. isn't everything at an instant?

4. Dec 5, 2006

### PhanthomJay

This problem is difficult using Newton only. You say you didn't learn energy conservation , but you do have some sort of formula relating to work and energy, so that means you have studied work energy methods but not the conservation of energy principle? Do you know the formula for PE of a spring??And regarding 'average' forces in a spring, when you use F = kx, the force is 0 when x is 0, then builds up to a maximum force when x is at its final displacement. The average force is kx/2. Thus, the work done by the spring is F_avg(x) = kx/2(x) = 1/2kx^2.

5. Dec 5, 2006

### PhanthomJay

Maybe too late, but check out my comments in red. I don't like doing the problem this way, but it works.

6. Dec 5, 2006

### dustpal

woah, that solved all the problems....like even my other ones similar to this one, you are a lifesaver...i think, idk, we'll see the results later. although...my other has me finding the x(max)...which i'm finding to be almost impossible again

Last edited: Dec 5, 2006