# Determine the speed of the rock as it leaves the spring.

• HumanPerson
In summary: So what do we do about it? (1) We can tell the question setter that they are wrong, but that doesn't help the student. (2) We can tell the student that the question setter is wrong, but that doesn't help them pass exams. (3) We can tell the student to ignore the fact that the question setter is wrong and do as they are told. I think (3) is what we have to do, but it rankles with me. Is it possible that any member of the site team can comment on this? Are we absolutely certain that the question setter is wrong? I can see that if, for example, the spring were winding and unwinding with a

#### HumanPerson

Here is the question:
A spring is compressed 20cm. When it is released, it exerts an average force of 57.2 N on a 7.3 kg rock, shooting it across a frictionless floor. Determine the speed of the rock as it leaves the spring.

## Homework Equations

W=change in energy
Work energy theorem
W=Fd

## The Attempt at a Solution

W=Fd
=(0.20cm)(57.2N)
=11.44J

W= change in EP elastic
=1/2kx^2
Rearrange formula to solve for k (spring constant)
=10.69579357 N/m

I think you're going the wrong direction with this. You have the force applied and the mass affected, so you can solve for the average acceleration of the rock (F=ma, a=7.85616438 m/s2). With the acceleration and Δx (20 cm or .2 m), you can apply the formula vf2 = vo2 + 2(a)(Δx). Since vo is 0, 2 times a times Δx = your final velocity squared.

HumanPerson said:
A spring is compressed 20cm. When it is released, it exerts an average force of 57.2 N
This is awkward.
Unless stated otherwise (e.g. as "average force with respect to distance") average force means the average over time. ##F_{avg} = \frac{\int F.dt}{\int dt} = \frac {\Delta p}{\Delta t}##, where p is momentum. On that basis, you need to use the fact that the spring will expand according to simple harmonic motion.
However, it's 10 to 1 that the problem setter has screwed up and expects you to take the average force as being an average over distance: ##F_{distance avg} = \frac{\int F.ds}{\int ds} = \frac {\Delta E}{\Delta s}##, so I'll answer below on that assumption.
HumanPerson said:
W=Fd
=(0.20cm)(57.2N)
=11.44J
So far so good, but you no longer care about the spring. You know how much work has been done on the rock. Deduce its speed from that.

These give you the same answer... Work in = Kinetic Energy Final, right?

HumanPerson said:
W= change in EP elastic
=1/2kx^2
Rearrange formula to solve for k (spring constant)
=10.69579357 N/m
This formula is not useful to you here, because you are told the spring equation is not that of a normal spring. In a normal spring, F=kx. But in this exercise, you are told to use F=constant, independent of x.

mrnike992 said:
These give you the same answer... Work in = Kinetic Energy Final, right?
That should work.

mrnike992 said:
These give you the same answer... Work in = Kinetic Energy Final, right?
What give the same answer? Are you saying that both definitions of average force give the same answer?
This is annoying.. I posted a long response to this last night and it has disappeared. Maybe some site monitor mistakenly thought I was providing you with a complete answer to the OP, so I'm not going to try typing it all again.
The upshot was that I demonstrated that using the correct definition of average force gives a different answer The ratio between the answers is sqrt(pi/4).

I solved it the way I showed in my original post and with the equation (Work = kinetic energy final) and both most certainly gave me the exact same answer, accurate to like twelve or fifteen sig-figs, however many my calculator allowed.

haruspex said:
using the correct definition of average force gives a different answer

Okay, I think I see what you're saying. But this is a high school physics question, I'm sure the question is implying a force over a distance, so it would be an Energy problem, as indicated by the formula for potential energy in a spring. It does say to find the speed as it leaves the spring, and the spring was compressed .2m, so I believe it would be indicating that the average force over the distance from the initial position to the end of the spring, .2m away.

mrnike992 said:
Okay, I think I see what you're saying. But this is a high school physics question, I'm sure the question is implying a force over a distance, so it would be an Energy problem, as indicated by the formula for potential energy in a spring. It does say to find the speed as it leaves the spring, and the spring was compressed .2m, so I believe it would be indicating that the average force over the distance from the initial position to the end of the spring, .2m away.
I agree that this is almost certainly what the problem setter intends here, but I hate the thought of students being taught that this is a valid interpretation of average force. It leads to the bizarre situation that mass * average acceleration does not always equal average force.

## 1. How is the speed of the rock calculated?

The speed of the rock can be calculated by dividing the distance the rock travels by the time it takes to travel that distance. This is known as the average speed.

## 2. What factors affect the speed of the rock?

The speed of the rock can be affected by the force used to launch the rock, the angle at which the rock is launched, and any external forces such as air resistance or friction.

## 3. What units are used to measure the speed of the rock?

The speed of the rock is typically measured in meters per second (m/s) or kilometers per hour (km/h). Other units such as feet per second (ft/s) or miles per hour (mph) may also be used.

## 4. How does the speed of the rock change as it leaves the spring?

The speed of the rock will gradually decrease as it travels through the air due to the effects of air resistance. However, if the launch is done in a vacuum, the speed of the rock will remain constant as there is no air resistance present.

## 5. Can the speed of the rock be calculated using other methods?

Yes, there are other methods to calculate the speed of the rock such as using the principles of conservation of energy or using the equations of motion. However, the most common and straightforward method is to use the distance and time traveled.