Solve Block Pulled Upward: Find Velocity in Terms of m, g, b

[Poetria]

Homework Statement

An block of mass m is at rest on a horizontal surface. At t=0, a vertical upward force is applied to the block (for example, by pulling upwards on a string attached to the blcok). The magnitude of the vertical force depends on time as F=bt where b is a constant coefficient. The acceleration due to gravity is g.

We strongly suggest you switch perspective, and for parts 4-6 use the instant the block starts accelerating as the time t=0.

Find the velocity v of the block when the block has upward acceleration of magnitude g. Answer in terms of m, g, and b.

Homework Equations

The Attempt at a Solution

Well, I thought it was simple but something isn't working.

Fnet=b*t-m*g
t_0 - time when acceleration starts = m*g/b
t - time when acceleration is g = 2*m*g/b

I integrated b*t-m*g with respect to time and evaluated it from m*g/b to 2*m*g/b.

Integral (m*g/b, 2*m*g/b) (bt-mg)dt

I got : (g^2*m^2)/(2*b)

It is not correct but I am at my wits' end.

 

[BvU]

The dimension of your answer is force times time , not the dimension of a velocity.

You didn't follow the strong suggestion in the exercise, which is your good right, of course: the answer shouldn't change.
If dividing by m isn't enough, you could still try it the easier way.
If you show your work, someone will be glad to comment/assist.

 

[Poetria]

Oh, ok, I think I understand what you mean. I will try. Many thanks.

 

[Poetria]

Yes, silly me. It is correct now. :)

 

[BvU]

Conclusion: you did the difficult version impeccably. Just forgot to go from F to a with a constant factor 1/m.
Not bad at all.

 

[Poetria]

Thank you very much :) but I have screwed something up in the next part :( :

How high h is the block above the horizontal surface at the instant when the block has the upward acceleration of magnitude 2g? Answer in terms of m, g, and b.

I started with velocity:

v=(b*t^2)/(2*m)

then I integrated it with respect to time to get distance:

integral (m*g/b, 3*m*g/b) (b*t^2/(2*m))dt

t=3*m*g/b when the net force is 2g.

2*g*m=bt-m*g
t=(3*g*m)/b

I got:

(13 *g^3 *m^2)/(3 *b^2)

Hm, I thought that there might be a problem with the initial distance but it is zero, isn't it? The guys here had a similar problem: https://www.physicsforums.com/threads/distance-from-non-uniform-linear-acceleration.736763/

 

[BvU]

v=(b*t^2)/(2*m) is taking off from zero at t=0. But the block doesn't take off until t = mg/b

Or does this mean you followed the perspective switch that was so pleasantly suggested ? :)

In that case you want to look at the integration boundaries again ...

[edit] And don't get distracted by the load of hogwash in thread 736763. For your problem simple analytical integrals (that you are already quite good at) will solve the thing flawlessly.

 

[Poetria]

Oh, I got it. Indeed integrals work. :) :)
I am wondering what I will screw up in the next. :(

Thank you very much for your patience. The only excuse I have is that my background is quite different ( I studied philosophy) and I am suffering all this pain out of love for Sir Isaac. :)

 

[BvU]

Well, never too late.
Newton was very special, indeed. I loved reading 'Dark matter' (Kerr) and the baroque trilogy (Stephenson).

 

[Poetria]

When I took an AI online course (run by Thrun) I met Stephenson's fan who recommended him to me. :) He is a great writer indeed.

There is a wonderful site dedicated to Newton: http://www.Newtonproject.sussex.ac.uk/prism.php?id=1

It is very difficult to interpret Newton. For example, I have read that very few mathematicians at the time (among them Jacob Hermann) noticed that it was possible to dig out the concept of kinetic energy out of his analysis in Propositio 41 (Principia). This was the same as Leibniz's vis viva with the difference that Leibniz was searching for a metaphysical principle in the universe. I hope it is never too late. :)