# Motion and Force: Dynamics

1. Sep 17, 2008

### CDink

Im having trouble with 10 problems due next wednesday...help guys...I was having a rough week and i don't understand any of this at all. so im gonna need even more help with these than last time

1.A skier traveling at 34.9 m/s encounters a 17.8 degree slope. If you could ignore friction, to the nearest meter, how far up the hill does he go?

?, ok i know W=mgSin(theta)...i don't know how to put all that in though im just so confused.

2.If the coefficient of kinetic friction in the previous problem was actually 0.12 and the slope 30 degrees, to the nearest meter how far up the hill does he go?

same.

3.You have a mass of 68 kg and are on a 58-degree slope hanging on to a cord with a breaking strength of 200 newtons. What must be the coefficient of static friction to 2 decimal places between you and the surface for you to be saved from the fire?

this one i really wanna think i have...like i can remember doing a similiar problem in class but im drawing a blank

4.In the previous problem if the coefficient of static friction is zero, to the nearest tenth of a degree, what would the incline angle have to be in order for the cord not to break?

Same.

5.The red box has a mass of 20.9 kg and the blue box has a mass of 12.1 kg and the force is 289 N. To the nearest tenth of a m/s2 what is the acceleration of the combination?

ok im thinking this one is set up like:

289=a(12.1+20.9)?

how on or off is that?

6.To the nearest newton in problem 5 what force does the blue box exert on the red box?

im lost on this one.

7.A child pulls a wagon with a force of 64 N by a handle making an angle of 11 degrees with the horizontal. If the wagon has a mass of 4.8 kg, to the nearest hundredth of a m/s2 what is the acceleration of the wagon?

in this one(as in most of these) i feel like i know what i have to do...only the formulas and stuff...im lost

8.To the nearest newton what would be the minimum force applied at that angle which would lift the wagon off the ground?

same.

9.A student stands on a scale in an elevator that is accelerating at 2.2 m/s2. If the student has a mass of 58 kg, to the nearest newton what is the scale reading?

i feel incredibly stupid i can't solve this one......

10.A 26 kg child sits on a 5 kg sled and slides down a 126 meter, 29 degree slope, to the nearest m/s what is his or her speed at the bottom?

Same with above.

I feel really dumb that i had to come with this many problems i have no idea how to solve...sorry guys.

2. Sep 17, 2008

### Dschumanji

You are spot on for number 5.

For number 6, how much force is needed to accelerate the blue block by the value you got in number 5? What is providing that force? What does newton's third law tell you about that force?

3. Sep 17, 2008

### CDink

ok, well the force on the blue block is 106.5N.....and the third law would mean that it's also giving a force of 106.5 on the red block?

4. Sep 17, 2008

### Dschumanji

Yup, that is correct

5. Sep 17, 2008

### CDink

stellar! well thats two down already!

6. Sep 17, 2008

### Dschumanji

You have the concepts down, but it seems your values are off by a bit and not rounded to the correct number of significant figures.

7. Sep 17, 2008

### CDink

ummm whered i make the mistakes...lets see.

289=a(12.1+20.9)
289=a(33)
289/33=a
a=8.758=8.8

theeeen

(I don't know what to put for sigma forces sooo)

sum of the forces=(8.8)*(12.1)
sum of the forces=106.480=106.5

where was i off?

8. Sep 17, 2008

### Dschumanji

According to sig fig rules, 33 should be written as 33.0 which gives you 3 significant figures. Dividing 289 (3 sig figs) by 33.0 means you should round your answer to 3 sig figs. So your answer for the acceleration should be 8.76.

12.1 multiplied by 8.76 means you round to 3 sig figs giving answer of only 106.

9. Sep 17, 2008

### CDink

right right..my teacher would have accepted that though..but i disregarded sigfig...not good i know.

well i spose i'll keep checking back for the rest of the help...thanks a lot for your help already.

Last edited: Sep 17, 2008
10. Sep 18, 2008

### CDink

ok...I think i have the elevator problem...

would it be 2.2+9.8=12.....*58=696N?

Last edited: Sep 18, 2008
11. Sep 18, 2008

### CDink

ok as far as 1 goes...all i've set up is the diagram and im stuck...i don't know the objexts mass...and i don't know the equation to use.

sorry bout the triple post...but i didn't want this to go to the next page and i wanted to show what i can do on the first one.

12. Sep 18, 2008

### Dschumanji

The question is rather vague in my opinion. Is it accelerating up or down at 2.2 m/s2? That determines whether or not his apparent weight is larger or smaller than his actual weight. If it is accelerating up then you're right.

13. Sep 18, 2008

### Dschumanji

You have the correct equation for the magnitude of the net force acting on the skier as he moves up the hill. What does Newton's second law tell you about the force acting on the skier? Answering that question should give you insight as to why the mass of the skier is not included in the problem statement.

What variables do you know the value of so far, what are you looking for? When the skier reaches his maximum height on the hill what will his velocity be? There is one constant acceleration formula that relates all the knowns to the unkowns.

14. Sep 20, 2008

### CDink

i need to find delta x....and all i know is the slope of the hill and the speed of the skier...and his velocity at the final will obviously be zero....thats all i can get..i know 2nd law but can't seem to understand it.

15. Sep 20, 2008

### Dschumanji

You have already shown that that magnitude of the net force acting on the skier when he is on the slope is equal to mgSin(theta). Newton's second law says that the net force on an object is equal to ma. In other words:

ma=mgSin(theta)

The m's cancel out leaving you with:

a=gSin(theta)

The acceleration of the skier does not depend on his mass.

You now have all the information you need to find the distance he goes up the slope.

Last edited: Sep 20, 2008
16. Sep 21, 2008

### CDink

ok, well acceleration=3.5 in that case...

and i think the kinematic equation i need is....x=26-0/2(3.5)? am i right in saying that...

so that means x=3.8, so he travels 3.8 meters?

17. Sep 21, 2008

### Dschumanji

Your value for the acceleration is wrong and your equation makes no sense. It shows x=26.

Use this formula:

v2=vi2+2a(x-x0)

Solve for (x-x0)

18. Sep 21, 2008

### CDink

ok i got -4.333...should i have used -9.8 when i found a? I'll assume yes

so that would make my a=-3.5.....so he traveled......3.7 m?

19. Sep 21, 2008

### Dschumanji

The skier is going nearly 126 km/h, do you really think he would only travel 3.7m before achieving a velocity of zero?

Can you start showing your work? I have no idea how you are getting such values when the magnitude of the acceleration is equal to

a=gSin(theta)

20. Sep 21, 2008

### CDink

ok, sorry.

a=gsin(theta)

a=-9.8sin21.2

a=-3.544

------------------------------

0-26/2(-3.544)=(x-x0)

-26/-7.088=(x-x0)

3.668=(x-x0)

theres my work wheres my mistake am i noticed you underlined magnitude...why? i honestly don't know whats different between that and actual acceleration.