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 [Broken]...
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...
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...
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...
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.
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...
"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...
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...
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...
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...
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...
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...
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...
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...
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...
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...