Homework Help: Linear Impulse and Momentum

1. Mar 5, 2015

_N3WTON_

A stop block, s, prevents a crate from sliding down a $\theta = 33 \hspace{2 mm} degrees$ incline. A tensile force $F = (F_{o}t) N$ acts on the crate parallel to the incline, where $F_{o} = 325 \frac{N}{s}$. If the coefficients of static and kinetic friction between the crate and the incline are $\mu_{s} = 0.325$ and $\mu_{k} = 0.225$, respectively, and the crate has mass of $m = 50.8 kg$, how long will it take until the crate reaches a velocity of $v = 2.44 \frac{m}{s}$ as it moves up the incline.

2. Relevant equations

3. The attempt at a solution
I know when $f < f_{max} = \mu_{s}mgcos(\theta)$, the system isn't moving, and the net force is:
$F - mgsin(\theta) - f = 0$.
$f = F-mgsin(\theta) < \mu_{s}mgcos(\theta)$
When $t = t_{1}$, $f = \mu_{s}mgcos(\theta)$ so I have:
$325t_{1} - mgsin(\theta) = \mu_{s}mgcos(\theta)$
$t_{1} = \frac{mg(sin(\theta)+\mu_{s}cos(\theta))}{325}$
Now plugging in my values:
$t_{1} = \frac{(50.8)(9.81)(sin(33)+(0.325)cos(33))}{325}$
$t_{1} = 1.25 s$
So, after 1.25 seconds, the block moves up along the incline and the net force is:
$F - mgsin(\theta) - \mu_{k}mgcos(\theta)$
The impulse of the net force is:
$\int_{t}^{t_{1}} (F-mgsin(\theta)-\mu_{k}cos(\theta)) = m(v-0)$
$\int_{t}^{t_{1}}(325t - mgsin(\theta)-\mu_{k}mgcos(\theta)) = mv$
$162.5(t^{2}-t_{1}^{2}) - mg(sin(\theta)-\mu_{k}cos(\theta))(t-t_{1}) = mv$
$162.5(t^{2}-(1.25)^{2})-177.4(t-1.25) = 123.95$
$162.5t^{2} - 177.4t - 126.76 = 0$
When I solve this quadratic, I get $t = 1.58 s$. However, this answer is not correct and I am not quite sure where I went wrong. Any help is appreciated. Thanks.

2. Mar 5, 2015

TSny

Does your answer represent the total time since t = 0? [Edit: Yes, it does. Your work looks good, but I haven't checked your numbers. Let me try it.]

3. Mar 5, 2015

_N3WTON_

Yes, I'm really not sure where I went wrong on this one...I'm thinking maybe I entered a value incorrectly but I can't see where. I was thinking that having another set of eyes look over the problem would really help :)

4. Mar 5, 2015

TSny

When you factored out -mg in the third to last line, did you mess up a sign?

5. Mar 5, 2015

_N3WTON_

I believe that I did, I had:
$-mgsin(\theta)-\mu_{k}mgcos(\theta)$
$-mg(sin(\theta)-\mu_{k}cos(\theta)$
$-mg(sin(\theta)+\mu_{k}cos(\theta))$ Thank you !

6. Mar 5, 2015

TSny

Yes. I hope it works out now.

7. Mar 5, 2015

_N3WTON_

I'm trying it now...

8. Mar 5, 2015

_N3WTON_

so after making the correction, I ended up with
$162.5t^{2} - 365.458t - 126.7645 = 0$
$t = 2.55 s$
This answer is incorrect, the correct answer according to Pearson is $t = 2.01 s$ :/

9. Mar 5, 2015

TSny

I get a different number in place of your -126.7645.

10. Mar 5, 2015

_N3WTON_

is it 78.96 or something close to that?

11. Mar 5, 2015

TSny

Yes.

12. Mar 5, 2015

_N3WTON_

Ok, I see where I went wrong now, thank you very much!

13. Mar 5, 2015

TSny

OK. Good.

14. Mar 5, 2015

_N3WTON_

and using that number I do get 2.01 seconds after using the quadratic equation :)

15. Mar 5, 2016

mohjee

Why do you have the time you found in the upper limit?

16. Mar 5, 2016

mohjee

Why do you have the time you found in the upper limit?

17. Mar 5, 2016

TSny

You are right, t1 should be the lower limit. When _N3WTON_ evaluated the integral, t1 was used as the lower limit. So, it's OK.

18. Mar 6, 2016

mohjee

Okay, thank you for that, was just making sure, I noticed that the numbers made sense.