# Distance traveled by a block due to unbalanced force

## 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.

Related Introductory Physics Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
Hello Romel, welcome to PF.
Please note that 'relevant equations' means something different than what you present there: more like Fnet = Fext - Ffriction, 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 Ffriction 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 ?

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 !

Last edited:
BvU
Homework Helper
2019 Award
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)