Distance traveled by a block due to unbalanced force

[Romel]

Homework Statement

A 60 N block in horizontal plane is pushed horizontally by force P = t^3 + 10t. The frictional resistance f = 6 - t^2. What is the distance traveled by the block after 2 seconds.

Homework Equations

The net force F = (t^3 + 10t) - (6 - t^2) = t^3 + t^2 + 10t - 6.

The Attempt at a Solution

There are actually three questions in this problem, acceleration, velocity and distance, all after 2 seconds. I encountered no problem with the acceleration which is from F = Ma
a = (g/W) (t^3 + t^2 + 10t - 6) and after 2 seconds, a = 2.398 m/s

Also no problem with the velocity by taking dv / dt = a, it is straightforward to integrate dv = a dt from known acceleration equation above. The equation of velocity is v = (g/W) [ (t^4 / 4) + (t^3 / 3) + 5t^2 - 6t ].

My question is the distance, with equation ds = v dt.

I have difficulty deciding which one to take for the initial time. Since at t = 0, the frictional resistance f is greater than the horizontal force P. I initially took t from zero to 2, but found that motion will start to impend not until 0.552 590 6474 seconds. Integration to distance equation will yield 0.7 for limits from 0 to 2 and 0.8 if limits is from 0.552 590 6474 seconds to 2 seconds. So which one is the right answer, I need your opinion. Thanks.

 

[BvU]

Hello Romel, welcome to PF.
Please note that 'relevant equations' means something different than what you present there: more like F_net = F_ext - F_friction, etcetera. But you do mention a few under 3.

I believe your $F = ma$ at t=2 (**). (As a physicist, I resent the author's wording 'A 60 N block' because that doesn't tell me the mass, but never mind).

I would be surprised if your integration to get v is correct. After all, the thing does not move for a while, and you integrate from 0 to 2.

You yourself only become suspicious when working on the displacement. Better late than never, I would say. But you should realize that if s is 0 for a while, then v = 0 for that while too!

And I just love the 0.552 590 6474 ! You sure it's not 0.552 590 6473 ? :) What I mean is that you can't present so many digits (the 60, the 2, the coefficients in P and F_friction are all single-digit). Best is here to give it a name like t₁, work with that (if you're lucky it might even cancel at some later moment -- not in this exercise) and fill in a value only at the very last step.

(**) But - just to be sure - can you show me how (28 - 2)(g/W) gives 2.4 ?

And I also get quite different v and s ! Are you working in some strange system of units ?

[edit]After a lot of work on my side, I see the mixup. Do yourself a favor and check what you are doing. If that's done and OK, check what you are posting !

 

[BvU]

There is another discussion lurking here: the exercise clearly states that 'The frictional resistance f = 6 - t^2'

This is very unsatisfactory. It would be dead wrong if it would say 'The frictional force f = 6 - t^2'

If I am lenient to the author, I expect he has written somewhere in the chapter something along the line that frictional resistance is the maximum friction force magnitude. But he (/she?) could have been less devious by writing $|f_{max}|$ instead of f.

(It is dead wrong in another respect: there are no units)