Need steps to help me solve this physics problem?
152421
Well I wont solve it for you \nbut I will help you. \n\nThese problems always involve the equation \ns=s0+v*t+(at^2)/2\n\nwhere s is displacement\nv is speed\nand a is acceleration\n\nNow you will always have acceleration of gravity g\n\nbut when the freak comes out of the canon, they will have a certain velocity. \n\nThey will be accelerated as the spring releases its energy, and then leave the barrel with a constant velocity, at that angle. \n\nThey will have a horizontal velocity and a vertical velocity. \n\nso you will need two equations\n\nx= Vx *t + .5*Ax*t^2\n\nand y=Vy*t+.5*Ay*t^2\n\nOne for vertical travel, and one for horizontal travel\n\nThe horizontal travel will will assume Ax=0 (not really true, but close enough)\n\nThe vertical travel, assume Ay=-g, acceleration of gravity. \n\nTo find the speed out of the barrel at the start, \n\nremember Hooke's law\n\nF=kx. \n\nSo you know k and you know x, so you know F\n\nYou also know Newton's 2nd law, F=ma.\n\nSo you know a =F/m\n\nNow you have to find the velocity V. \n\nAs the spring releases, it accelerates the freak with acceleration a. The only question is, over what distance is this acceleration acting? \n\nWell one idea is that it acts over the compressed distance. Another idea is that it acts over 2 times the compressed distance, since it will oscillate in approximately simple harmonic motion, and will continue to push that entire distance. I would guess the latter.\n\nSo put this in \n\ns=.5*at^2, \nand compute t\n\nFrom t, you can figure out the speed at launch: \n\nV=a*t\n\nSo you have the initial velocity V, which you can resolve into vertical and horizontal components, and then figure out how long it takes the freak to return back to the ground, and in that time, how far will the freak travel horizontally. \n\ngood luck\n\n- - - - - - - - - -\n\nkirchwey is correct. The same force is not applied through the entire distance that the spring oscillates. The force F=kx is the maximum force, only at the beginning of the freak's acceleration. I was thinking about that so I returned to this problem (it has been a long time since i did these problems). \n\nThe simple way to address this is to assume that all of the potential energy of the spring .5*k*x^2 is imparted to the freak, and turned into the freak's kinetic energy, .5m*v^2. \n\nAs kirchwey points out, there might very well be a new equilibrium position for the spring due to the freak's weight, or at least the component of the freak's weight that is projected along the axis of the tube. If the spring was not at all compressed by anything but the freak's weight, then if the spring was "released" the freak would not go anywhere. What one has to decide is, is the equilibrium point the problem mentions the equilibrium point for an unloaded spring, or the spring with freak compressing it a bit. Whatever you decide will determine how you solve this problem. I would try it both ways and see what the different answers are (they should not be too different because the freak does not weigh that much, relative to the spring force).
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
610467
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
694807
Use conservation of impulse and conservation of kinetic energy.\n\nmv + 0 = mu + mw\n(1/2)mv2 = (1/2)mu2 + (1/2)mw2\n\nu + w = 1 (consider it as a straight line in a graph)\nu2 + w2 = 1 (consider it as a circle in that graph)\nWhen you solve this set of two equations, you get\nu = 0 and w = 1 (m/s)\n(Find the useful intersection. The other one is is the initial situation.)\n\nTh
855600
u can also solve this by takin s= -100m\nn by the formula S= ut + 1/2 gt^2\ng = 9.8m/s^2. u = -6m/s .......4 the 2nd part\nand \n\nfind time of the 1st half by the formula \n=> v = u + at....[ where u = 6m/s, v = -6m/s(velocity in oposit direction)]\n\nALSO REMEMBER THAT WEN A TRAJECTORY AGAIN REACHES ITS STARTIN PT.  IT HAS THE SAME MAGNITUDE OF VELOCITY\n\n-ve sign is given to all dat is below the level or in the direction below the startin point\ncombine these 2................ n voila..\nwith time u can master this easy.... n logical method
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
909402
Well, it's Wednesday morning.  I hope this answer still has some value to you.\n\nI started by summing the torques at the hinge.  There are three:\nT(light)=5*9.8*7.5*cos(37)\nT(cable)=-tension*6.31*sin(37)\nT(pole)=8*97.5/2)*cos(37)\n\nThese all sum to zero.\n\nI computed tension=140.5\n\nThe sum of the horizontal forces must be zero.\nThese are the tension in the cable and the horizontal reaction force at the hing.  So the horizontal component is 140.5\nThe vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge\nthe vertical component is (8+5)*9.8\n=127.4\n\nI checked the answer by computing the magnitude of the reaction force two ways.  Using Pythagorean theorem \nsqrt(140.5^2+127.4^2)\n=189\n\nI then summed the forces parallel to the pole:\n5*9.8*sin(37)\n8*9.8*sin(37)\n140.5*cos(37)\n=189\n\nj
796334
Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline\n\nSFx = m*a = m*g*sin(q) - Fs\n\nwhere m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque\n\nST = I*a/r = r*Fs\n\nso that we have for the statice friction force\n\nFs = I*a/r^2.\n\nPlugging this into the first equation\n\nm*a = m*g*sin(q) - I*a/r^2\n\nand solving for the acceleration\n\na = m*g*sin(q)/(m + I/r^2)\n\nNow, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is\n\nL = 0.5*a*t^2\n\nso that\n\nt = sqrt(2*L/a)\n\nAlso, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by\n\n2*pi*L/(2*pi*r) = 0.5*alpha*t^2\n\nwhere alpha is the angular acceleration and is equal to a/r. \n\nConservation of energy can be employed as well\n\nm*g*h = 0.5*m*v^2 + I*w^2\n\nwhere h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields\n\nv = sqrt(2*m*g*L*sin(q)/(m + I/r^2))\n\nwhich is equal to the acceleration times the time traveled. From this \n\nt = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)