B Lifting an object with a mass difference of just 1kg

[nuckshuck11]

Hello, can anyone help me out with this?

If i want to lift an object that weights 100kg and I am going to do it with a 101kg of force (990,81 N ).
Then my acceleration would be;
a = F/m
a= 990,81N / 101kg = 9,9081 m/s2
but gravity is pulling it down so then I am acually accelerating it with 9,9081 - 9,81 = 0,0981 m/s2

SO, now that i know I am accelerating my object 0,0981 m/s2, does it mean it gets faster every second?
Like is it getting faster and faster every second that force is applied with only 1 kg more than its own weight? 0,0981+ 0,0981
+0,0981 ect..??

Sorry, I am feeling dumb but can't find any answer online

[Moderator's note: approved in a technical forum for its principle nature, rather than a homework question.]

 

[Doc Al]

Then my acceleration would be;
a = F/m

Note that "F" here should be the net force on the object. Better to write it as a =∑F/m.

SO, now that i know I am accelerating my object 0,0981 m/s2, does it mean it gets faster every second?

Sure. There's a net force on the mass and thus an acceleration.

 

[Richard R Richard]

Two forces act on the object, the weight equal to $W=m\cdot g = 100kg 9.81m /s^2$ and the force to lift it, which you say is $F_l=101kg 9.81m/s^2$

Newton's second law tells you that the vector sum of all the forces is equal to the resulting acceleration times the accelerated mass.

$$\displaystyle \sum F = F_l-W = ma$$

$$101kg 9.81m / s^2-100kg9.81m / s^2= 100kg \cdot a$$

$$\displaystyle a = \dfrac {\sum F} {m} = 0.01\cdot g = 0.0981m / s^2$$

 

[kuruman]

Sorry, I am feeling dumb but can't find any answer online

Maybe you're feeling dumb, but your original post indicates that you figured it out all by yourself and correctly. That's not dumb in anyone's book, yours included. Who needs online answers when one has a brain and can put it to work?

 

[bob012345]

Hello, can anyone help me out with this?

If i want to lift an object that weights 100kg and I am going to do it with a 101kg of force (990,81 N ).
Then my acceleration would be;
a = F/m
a= 990,81N / 101kg = 9,9081 m/s2
but gravity is pulling it down so then I am acually accelerating it with 9,9081 - 9,81 = 0,0981 m/s2

SO, now that i know I am accelerating my object 0,0981 m/s2, does it mean it gets faster every second?
Like is it getting faster and faster every second that force is applied with only 1 kg more than its own weight? 0,0981+ 0,0981
+0,0981 ect..??

Sorry, I am feeling dumb but can't find any answer online

[Moderator's note: approved in a technical forum for its principle nature, rather than a homework question.]

It seems to me that you are trying to grapple with the concept of acceleration and how a small force difference can lead to a large change in velocity over time.

First, it's a bad habit to say an object weighs 100 kg. That's its mass not its weight. It's weight is 981N if we are on Earth and g is $9.81 m/s^2$. It would weigh less on Mars but still be 100 kg.

Next, to move that object you wanted to apply a greater force than its weight. You chose a force of 990.81N and thus the net force is 9.81N which is only 1% of the weight of the object and the acceleration is $0.0981 m/s^2$. That means the velocity of the object in a straight line pointing up in this case increases by $0.0981 m/s$ each second. The relationship is linear in time and goes simply as velocity = acceleration X time for a constant acceleration.

Actually, the formula is $v = v_0 + a ⋅ t$ where $v_0$ is the initial velocity if there is one. You have a small net force and a small acceleration but if you apply it for a longer time, say 1000 seconds, you end up going at $98.1 m/s$. Likewise, a larger acceleration, say $9.81 m/s^2$ will get to the same speed in just ten seconds. Play around with the formula if you like to get a feel for it.

As an aside, an acceleration of only 1% of g ($0.0981 m/s^2$) would be very useful for space probes. For an exercise, calculate how fast would the probe be going in a day, a week, a month or a year?

 

[Nugatory]

A smallish point:
You can save yourself some grief and make life easier for your audience by taking $g$ equal to 10 m/sec^2 instead of 9.81 when you don’t need better than 2% accuracy.

 

[anorlunda]

A frequent source of confusion is that we live in a world dominated by friction. In that environment, the faster things move the harder you have to push to overcome the friction. In 322 BC, Aristotle believed that F=mv was the correct equation. Not until the year AD1687 did Newton publish the F=ma version.

But Newton's laws use F to mean the sum of all forces including friction. In space, object move with zero friction. So the planets orbit indefinitely. But space is not the ordinary experience of those of us who are not astronauts. In space, Newton's laws are much easier to test and to visualize.