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1. Simple Pulley Problem

Ah thank you, that's what I forgot. I'm having issues with this other problem "A hoop is released from rest at the top of a plane inclined at 16 above horizontal. How long does it take the hoop to roll 16.4 m down the plane?" Another energy problem. I believe the equation is...
2. Simple Pulley Problem

That was quickly written, I meant to have those masses be different. -F_fk = 1/2m_1v^2-m_2gh This equation equates to 1.4m/s which is the correct answer while... -F_fk = 1/2m_1v^2+1/2Iw^2-m_2gh this equation which I believe is the correct equation equates to 1.13m/s. What am I doing...
3. Simple Pulley Problem

Homework Statement In Fig. 9.2, two blocks, of masses 2 kg and 3 kg, are connected by a light string that passes over a pulley of moment of inertia 0.004 kgm^2 and radius 5 cm. The coefficient of friction for the table top is 0.30. The blocks are released from rest. Using energy methods, one...
4. Finding Velocity(Momentum related)

Homework Statement Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose...
5. Potential Energy and Distance Graph

Homework Statement A particle moves along the x-axis while acted on by a single conservative force parallel to the x. axis. The force corresponds to the potential-energy function graphed in Fig. 7.45. The particle is released from rest at point A. [PLAIN]http://k.min.us/ijze8K.png [Broken]...
6. Work energy question. Rock down a hill.

I think I got it. Can someone check this? So my potential spring energy is incorrect I believe. In this problem the mass would be stretching the spring out to x length and not d length so the equation should look like this. mgx = 1/2kx^2 Since k = mg/d I can sub that in. mgx =...
7. Work energy question. Rock down a hill.

Work energy question. Fish on a spring. Question has been changed as I figured it out. New one is on fish and springs Homework Statement If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount d. If the...
8. Finding Velocity with Work-Energy Theorum

Okay, so the work friction does is u_k*m*g*s which would be -29.4J as the force is going the opposite of the displacement. Then the second block does mg*s which is 88.2J of work in the positive direction. Together that gives me a network of 58.8J in the positive direction. Now I can equal...
9. Finding Velocity with Work-Energy Theorum

Ah I see, thanks. I have another question I'm having issues with There is an 8 kg block on a table with a 6 kg block hanging off of a pulley on the end of a table. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the...
10. Finding Velocity with Work-Energy Theorum

Homework Statement There is an 8 kg block on a table with a 6 kg block hanging off of a pulley on the end of a table. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is 0.250 The blocks are...
11. Finding force

Would this just mean that the force of water resistance is equal to the force of the ship?
12. Finding force

Alright so couldn't I just find the total from what I did above, then dividing the power by the velocity to get force and then multiple that by .3 to get teh water resistance? 70% is going forward while the other 30% is water resistance? EDIT: Hmm that not correct. I really don't know what...
13. Finding force

Homework Statement The aircraft carrier John F. Kennedy has mass 7.4*10^7kg. When its engines are developing their full power of 280000 hp, the John F. Kennedy travels at its top speed of 35 knots. If 70% of the power output of the engines is applied to pushing the ship through the water...
14. Connected Blocks on a Ramp

Homework Statement Two blocks with masses 4.00 kg and 8.00 kg are connected by a string and side down a 30.0 inclined plane. The coefficient of kinetic friction between the 4.00kg block and the plane is .25; tha between the 8.00kg block and the plane is .35. (a) Calculate the acceleration of...
15. Kinematics - Boat over a constant River.

Well I found out what I needed to do to solve for theta. The Law of Sines made it easy to work it out. After that it was a piece of cake.
16. Kinematics - Boat over a constant River.

Yeah, had it been B this wouldn't be much of an issue. So the angle is going to be somewhere between 6.5 and 6.6 degrees. This was such a pain. Hopefully this type of question isn't on the exam. I'm doing the questions out of an e-book so maybe it's a typo on there.
17. Kinematics - Boat over a constant River.

Think I'm off in my math. I did as you said and got theta = 6.53 V_0y = 99.35 V_0x = -11.06+30 t = 4.02 When I check for the X length I'm a little high, at 76.26m instead of 75, so maybe I'll trying that again with more precision. So, is there no easier way to do this? This...
18. Kinematics - Boat over a constant River.

"100 sin(θ) = (400/75) (100 cos(θ) + 30)" Shouldn't this be 100sin(θ) = (75/400)100cos(θ) - 30? 100sin(θ) Would be the x component no? When I was first doing this I got it down to this but I had no idea how to get θ by itself and that's pretty much why I posted here. It seems like this...
19. Kinematics - Boat over a constant River.

Okay I think I see what you're saying. Ugh this is a tough question....
20. Kinematics - Boat over a constant River.

And that makes sense. The boat going straight across went some 120ish meters out, so the boat would need to point in a negative direction to get to C. That would mean that the 11.6m/min the boat is travelling west BUT the 30m/min river causes it to go 18.4m/min East which means it would make...
21. Kinematics - Boat over a constant River.

How did you get 11.57? I found 18.4 by using theta = 10.6 then finding the x component. I then assumed that that included the river's flow and that if I subtracted 30 I could have the boat's actual component which would be -11.6 which sounds right as the boat would need to be going against the...
22. Kinematics - Boat over a constant River.

Okay so what would we change to the equation to add in river flow? Would it be 75/400 = (100sin(theta) + 30)/100cos(theta) ?
23. Kinematics - Boat over a constant River.

I think you did 75/100 which is .75 and not 75/400 which is .1875. I'm still trying to grasp this. Theta would be 10.6 degrees but wouldn't that only work if the river wasn't flowing? Don't I need to add the river's 30m/min or is that included? The way it's set up now looks like the boat is...
24. Kinematics - Boat over a constant River.

I'm actually trying to reach point C, not point B. To reach point B I understand that I would need an x component of -30m/min to cancel out the river, but I don't know what I would need to get to point C. I know that it will have distances of y component 400m and x component 75m so I could...
25. Kinematics - Boat over a constant River.

Homework Statement A 400m wide river flows from west to east at 30 m/min. Your boat moves at 100m.min relative to the water no matter which direction you point. To cross the river, you start from a dock at point A on the south bank. There is a boat landing directly opposite at point B on...
26. Kinematics confusion

Yes I'm sure. It was multiple choice and apparently the answer is a negative...
27. Kinematics confusion

Homework Statement A motorist traveling at a constant speed of 160 km/h in a 50-km/h speed zone passes a parked police car. Three seconds after the car passes, the police car starts off in pursuit. The policeman accelerates at 2m/s^2 up to a speed of 30 m/s, and then continues at this speed...
28. Kinematics - Sled

Alright I think I got it now. Since I don't know the time it took to get to 14.4m I can't find an exact v_1 using an averaged velocity as it isn't't constant so for that I would use the original equation finding v_1 with the constant acceleration, time and distance which would give me an exact...
29. Kinematics - Sled

Homework Statement A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top. a) What is...
30. Kinematics - Drag Racer

Okay, so x_0 is the distance between the car and the dragster when the dragster takes off correct? So the last instant would suggest that x_0 is as small as it can be without the car crashing into the dragster. I need to find a t value(t-max) such that x_0 is as small as possible. That's the...