# Homework Help: Force applied on a falling mass

1. Mar 30, 2010

1. The problem statement, all variables and given/known data

Hey guys how are you all.

Lets see, I'm not a physicist (mathematician actually) but I'm involved in this discussion (some may be aware...) and would like some physics help. (This is not an assessment, tutorial, university, school etc question.)

Here's the scenario. (Hope I word everything right)

An object with a weight of 100N has a force applied to it so that it takes 4 seconds to fall 0.5m at a constant speed.

We then repeat except...
The same object then has a force applied to it so that it takes 0.5 seconds to fall 0.5m at a constant speed.

How do I calculate the forces applied (I guess they would be applied upwards) on the object in both scenarios?

To save you time... An object free falling 0.5m takes 0.3192754285 seconds.

Thanks guys!

2. Mar 30, 2010

### collinsmark

Welcome to Physics Forums!

In both cases (ignoring the free-fall case), the acceleration is zero, since the velocity is constant. That means that the force in both cases is identical. And, the applied force is equal in magnitude to the gravitational force, mg = 100 N (but opposite in direction).

The work done is also equal in both cases. Here,
W = Fz,
where F = mg = 100 N, and z = 0.5 m.
Using these units you can calculate the work in Joules.

The difference between the two situations is power (measured in Watts or horsepower, etc.)

Power is a measure of energy (work in this case) per unit time.

P = W/$$\Delta t$$

(If W is measured in Joules, and $$\Delta t$$ in seconds, then the average power P is in units of Watts)
Since each scenario performs the same amount of work in different amounts of time, the power is different in each scenario.

[The above ignores the brief moments of acceleration when the object first starts moving and stops moving. The problem is phrased such that the object is only considered when moving at a constant velocity. Likewise, the above response only considers the object when it is moving at a constant velocity.]

3. Mar 30, 2010

### Brilliant

Well, there a couple issues. First, you must know that for an object to fall at a constant speed the forces must be in equilibrium. So, if you're ignoring drag, then the upward force that needs to be applied is simply mass times the acceleration of gravity: mg

With Newton's second law:
$$\sum F = F_{up} + F_{g} = ma = 0$$
So,
$$F_{up} = F_{g} = mg$$

EDIT: What he said.

4. Mar 30, 2010

I see very interesting! Thanks for the reply.

This is based on a thread we are all arguing about on another site about the difference in force between fast repetitions and slow repetitions when lifting a weight. You may be aware of someone called wayne (Possibly waynelucky)? I believe he posted a thread of two here however his way of explaining isn't that clear...

His general consensus was the following.

Doing 6 reps (repetitions) taking 0.5 seconds on the positive part of the rep and 0.5 seconds on the negative part of the rep was better than doing 1 rep which had a 2 second positive and a 4 second negative.

Here, positive means... I suppose the lifting part, negative being the lowering of the weight.

Both scenarios last 6 seconds but the questions are does one scenario require you to use more force? Does this mean more muscular work and does this mean it will give you better results?

Now my thoughts (that I had just now, haven't been arguing this) was that you had to apply more force to lower a weight slower. However I can see that I was wrong and my understanding of physics isn't really that good. That was why I posted this question and could check before I began arguing rubbish.

Do you have any opinion on the rep speed argument? If the weight was say, 80lbs and the most you could lift was 100lbs, 6 reps at 0.5/0.5 vs 1 rep at 2/4..?

I would say that the complexities of a human movement like a 'rep' can't be broken down into a simple equation, like saying time = this, hence force = this. However the thing to note is that even when the acceleration is 0 the weight will still be applying tension on the muscles involved in the lift. It just remains to determine how much!