Force applied on a falling mass

AI Thread Summary
In the discussion, a scenario is presented where a 100N object falls 0.5m at constant speeds of 4 seconds and 0.5 seconds, prompting questions about the forces applied. It is established that since the object moves at constant velocity, the upward force equals the gravitational force (100N), regardless of the falling time. The key difference between the two scenarios lies in the power output, as power is defined as work done over time. Additionally, the conversation shifts to the implications of lifting weights at different speeds, debating whether faster repetitions require more force and lead to better results. Ultimately, the complexities of human movement and muscle tension during lifts are emphasized, indicating that physics cannot simplify the nuances of exercise performance.
Deadstar
Messages
99
Reaction score
0

Homework Statement



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!
 
Physics news on Phys.org
Hello Deadstar,

Welcome to Physics Forums!

Deadstar said:

Homework Statement



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!

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.]
 
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:
<br /> \sum F = F_{up} + F_{g} = ma = 0<br />
So,
<br /> F_{up} = F_{g} = mg<br />

EDIT: What he said.
 
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top