Recent content by Exuro89

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...

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...

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...

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...

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 (a)...

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 =...

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...

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...

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...

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...

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

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...

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...

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.

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.