[Cryptic™]

A 300 N mountain goat is standing on a mountain side inclined at 70 degrees. What is the frictional force that is keeping the mountain goat at rest.

Can you show me the diagram for this question please. I can do the rest.

[360]

A force F = (6i-2j) acts on a particle that undergoes a displacement of r=(3+j) metres. Find (a) the work done by the force on the particle and (b) the angle between F and r.

I've already done part (a) and found the work done to be 16J, just need to know how to find the angle between the vectors.

[SIM-55606]

A force F = (6i-2j) acts on a particle that undergoes a displacement of r=(3+j) metres. Find (a) the work done by the force on the particle and (b) the angle between F and r.

I've already done part (a) and found the work done to be 16J, just need to know how to find the angle between the vectors.

I assume you mean 3i?

Difference = r-F = (3i+j)-(6i-2j)=(-3i+3j) or (-3,+3)
tan^-1(Rise/Run) = Angle.
tan^-1(-1) = angle = 45(o)

[360]

I assume you mean 3i?

Difference = r-F = (3i+j)-(6i-2j)=(-3i+3j) or (-3,+3)
tan^-1(Rise/Run) = Angle.
tan^-1(-1) = angle = 45(o)

Cheers for that mate, seems straightforward enough

[img]http://www.webassign.net/sercp/p5-55alt.gif[/img]

If I want to find the work done in a graph like this, do I simply create a triangle under the line and find the area?

And if I wanted to find the total work done from x(initial) to x(final), do I add the areas above and under the x axis to find the work done (so I am actually subtracting the area under the x axis)?

[Feynman]

Cheers for that mate, seems straightforward enough

[img]http://www.webassign.net/sercp/p5-55alt.gif[/img]

If I want to find the work done in a graph like this, do I simply create a triangle under the line and find the area?

And if I wanted to find the total work done from x(initial) to x(final), do I add the areas above and under the x axis to find the work done (so I am actually subtracting the area under the x axis)?

Integrate it with respect to x. You could use either numerical integration methods, or just find the areas of the triangles.

[360]

No worries, got that. Thanks.

Some more for you guys, if you don't mind;

-Question 1-
A 7.80 g bullet moving at 575 m/s penetrates a tree trunk to a depth of 5.5 cm. a) use work and energy considerations to find the average frictional force that stops the bullet. b) assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

-Question 2-
A 0.600-kg particle has a speed of 2.00m/s at point A and kinetic energy of 7.50 J at point B. What is (a) its kinetic energy at A? (b) Its speed at B? © the total work done on the particle as it moves from A to B? (Part C is the part that is bothering me here, not so much (a) or (b). Do I just take the average force (force at point A and force at point B and divide by 2? So I can use that force in the W=FxD equation?)

-Question 3-
A 2100 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5 m before coming into contact with the top of the beam. Then it driver the beam 12 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

[OversouL]

Thank you for this. I think this is a perfect timing since I have physics class. I'm in my senior high right now. Also I'm very bad at math. I'm really happy about this thread, I hope you can help me soon, but Our class just started 2 days ago, maybe you can help me advance study when I have my books. ^_^

Bookmarked.

[Feynman]

F*d = E → F = E/d = mv²/(2d) = (.00780kg)(575m/s)²/(.11m) = 2.34E4 N

I = F t (I = change in momentum = p = mv, since final momentum = 0)
t = p/F = mv/F = (.0078kg)(575m/s)/2.34E4N = 1.91E-4 s

I'll work on 2 & 3 in about an hour. I have to run errands.

[SIM-55606]

Using the relationship between work and energy that
W = (delta)Ke-(delta)Pe

Therefore:

---

Part 3.

[360]

A backyard swimming pool with a circular base of diameter 6.00 m is filled to depth 1.50 m.

A) Find the absolute pressure at the bottom of the pool. (Done, but including it so as to understand the entire scope of the question). Answer is 115.7 kPa

B) Two persons with combined mass 150 kg enter the pool and float quietly there. No water overflows. Find the pressure increase at the bottom of the pool after they enter the pool and float.

No idea where to start with part B, since to use p=m/v I need the volume, but since I need h in the volume (which has changed) I'm not really sure.

[Feynman]

There's actually a much easier way to do it!

P = F/A = mg/A = (150)(9.81)/(9π)

Your answer is the increase in pressure, so the new absolute pressure is your answer from (a) + (b).

[360]

There's actually a much easier way to do it!

P = F/A = mg/A = (150)(9.81)/(9π)

Your answer is the increase in pressure, so the new absolute pressure is your answer from (a) + (b).

Makes a bit more sense...but where did you get 9pi from?

[Feynman]

P(0) = 101kPa; (ro) =1000kg/m^3 h1=1.5m
P = P(0) + (ro)gh1
p = 101000 + 1000*9.81*1.5
P = 115.7kPa

The next one you need to figure out how much water they displace. and find the change in h due to the displaced water.
To float they must displace the same mass of water as themselves.
So 150kg, they displace 150kg of water. 150kg of water is roughly .15m^3 (density = 1000kg/m^3)
so if v = A(base) h -> .15 = 9(pi) h2
h2 = .15/9(pi)
h2= 5.3052*10^-3m

So delatP = P(0) + (ro)*g*(h2) = 52.04 Pa

Don't know what Feynman was going on about?

Works out to be the same. You learn a lot of tricks and shortcuts working with it for a while.

Anyway, closed for the time being. If anybody needs help, PM me directly.

[Feynman]

Re-opening this. I'm not nearly as flexible with my availability as I used to be.

Again, I ask people to please read my rules and conditions.

[Comedic]

Please explain this to me if you can

Spoiler (Click to View)

Thank you so much for your help!