I figured out what I did wrong in this situation... cos-1(0) could be either pi/2 or -pi/2, and because the motion is coming to the end of a complete cycle I should have used -pi/2.
To answer your question, I determined the phase shift=phi by solving the position equation. I knew at t=0...
At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the...
Find the area bounded between the two curves
y=34ln(x) and y=xln(x)
Integration by parts: \intudv= uv-\intvdu
The Attempt at a Solution
First I found the intersection points of the two equation to set the upper and lower bounds. The lower...
I think the mistake you are making is in the formula that you are integrating. It seems like you were thinking washers while using the shell equation.
The general equation to use for the shell method is
2\pi\int R dx(dx can change depending on which variable you are integrating with respect...
because the velocity is downward it is negative, as well as the acceleration. you can't interchange them. it is just important to note the direction of the velocity depending on the equation you decide to use!
This equation is fine to use, but you will have to solve for acceleration (you have the initial and final velocities, plus the time!) Once you solve for acceleration, you can solve for the displacement over time.
k=spring constant so it is accounted for. the negative sign is correct, but you also have to take into consideration that you will be taking the square root of the magnitude of the work. the negative simply implies direction--the spring is being compressed.
okay here are my thoughts...
you were right in thinking that this is a conservation of energy problem, or at least that is what I did too. we know that the work done by the spring is equal to 1/2kx2, and we also know that work is defined as force times distance. you can solve for the force of...
Okay! well, if the exponents have the same base (ie 2^x-1=2^6-x) then you can set the exponents equal to each other. For your problems, you will have to manipulate the bases to get them to be the same... for example if you have 2^x-4=8, that also equals 2^x-4=2^3, so then x-4=3.
That seems to be just what my problem is... I can't see the relationship between b and h.
I tried setting up a triangle and solving for what b is equal to in terms of h but I am still getting the wrong answer.
I used cos(theta)=(1/2b)/h, differentiated, and solved for what db/dt was equal to.
A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 7 inches deep?
Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with diameter 54 m.
Surface area of a hemishpere=2pi*r^2
The Attempt at a Solution
A student has two ramps, both of which are at an angle of 30o. Ramp 1 is frictionless and ramp 2 has friction. The student also has two blocks, one for each ramp. She pushes the blocks up the ramps with the same initial velocity. The block on ramp 2 only travels a...