# 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.