Problem on oscillating on a horizontal frictionless surface, help me physics people!!!?
919525
Vibration:  ω = √(k/m) = √(100/4) = 5.0 rad/sec\n\nV at neutral point = Vo = Aω = 2*5.0 = 10 m/sec\n\nx-direction momentum: This momentum will be the same before and after dropping the 6 on the 4 →\n6*0 + 4*5 = (6+4)*v →  v = 20/10 = 2 m/s\n\nThe new KE = .5mv² = .5*10*2² = 20 Nm ; This will equal the new spring energy:\n\n20Nm = .5kA² → A = √(40/100) = .6324 m, so the change in amplitude is\na)  2-.6324 = 1.3676 m\n\nb) New period: T = 2π√(m/k) = 2π√(10/100) = 1.987 sec\n....Old period:  To = 2π√(4/100) = 1.2566 sec\n....Difference = .73 sec greater\n\nc)  KE1 = .5*4*10² = 200 Nm.  From (above) KE2 = 20 Nm\n....so, energy change is -180 Nm\n\nd) The energy loss is released as heat between the (momentarily) sliding surfaces of the 2 objects when they contact.
906380
First problem: You have assumed all the mass of the meter stick is concentrated at the lower end; obviously, it is uniformly distributed along the length of the stick.  Assume that the mass is concentrated at the center; then, the length of the oscillator will be 0.865 meters, and the period should correspond.  There are some complications involving moment of inertia which I am ignoring.\nSecond problem: You may be solving for the wrong thing.  It looks like you are solving for the frequency of oscillation, rather than the maximum speed of motion.  Some careful work with units will help show you what is going on.
479843
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
813323
Well, whilst she is sliding, ignoring air resistance, the only force acting on her is the force of friction. That's the most important part of this problem. As such, you know that Fnet = Ffriction. Ffriction = Fnormal (Uk), so you can solve for Ffriction. That's Fnet, and Fnet = ma. You know Fnet and her weight, so divide her weight by 9.8 m/s to find her mass. Solve that equation for a, and then plug it into one of the handy-dandy kinematic equations. :)
772547
Increase your elevation or find a court with less gravitational force.   It's  a physics thing.
799229
a. Calculate the horizontal component of the 485 N.\nFh = 485 N*cos35\nCalculate the force of friction.\nFf = 0.57*319 N\nCalculate the net force.\nFn = Fh-Ff\nCalculate the mass of the box.\nW = m*g\nCalculate the acceleration.\nFn = m*a\nCalculate time to move s = 4.00 meters\ns = (1/2)*a*t^2\n\nb. Repeat with uk = 0.75
212531
Finally a way for me to pick up that 7-10 split!  :)\n\nI agree with the basic premise of oscillating assuming a vacuum (since we are assuming no molten core, why not a vacuum as well?).\n\nBut, depending on the direction of the hole in relationship to the rotational axis of the earth, wouldn't the ball start bouncing against the sides of the hole due to the mechanics similar to that of a Foucault Pendulum?
291004
if an object is floating, the buoyant force on it equals its weight -- the vertical forces have to sum to zero.  try drawing a free body diagram to convince yourself of this.
698608
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
772799
Ok i'll help with your homework. Even though you may want to try a little harder.\n\nIt's not that difficult, you just have to remember that the oscillations ideally are represented by a sine or cosine function.\n\nlet's say    x = f (t)\n\nx is position and t is time. so we know f involves a 'sin' or 'cos'.\n\nThen you have to increase the amplitude of the wave to reach the numbers 55 and -55 (multiplying the sin or cos by some factor).\n\nYou can also notice that the sin or cosine wave doesn't start at time t=0. Therefore you may have to use t+something instead of just the variable "t".\n\nThe answer that involves the sine function is:\n\nx = 55 sin ( (π/30)*(t+10) )\n\nFigure out the calculations for yourself. You might even try finding the OTHER answer which involves the cosine.
744947
First you should draw a free-body diagram to understand the forces acting on the block.  In this case, you have 4 forces:\n\n1) 40 N force moving in the +x direction, let's denote it F_1\n2) frictional force moving in the -x direction, let's denote it F_2\n3) Weight, moving in the -y direction, let's denote it W\n4) Normal force opposing weight, let's denote it N\n\nNow you can write Newton's 2nd law equations:\n\nx-direction: F_1 - F_2 = m*a\ny-direction N - W = 0 (equal to zero because block does not move in this direction)\n\nUsing these equations, you know that N = W = m*g\n\nUsing F_2 = u*N and a = 6 m/s^2m, we can solve for thd desired variables:\n\na)\nF_2 = F_1 - m*a  ---------> 40 - (5)(6) ---------> F_2 = 10 N\n\nb)\nF_2 = u*N = u*m*g, now solve for u:\n\nu = F_2/(m*g) = (10)/(5*9.8) -----------> u = 0.204\n\n--------------\n\nHope this helps
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
872212
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)