Physics Question Spring load #15?
817632
c: 8cm.\n\njust look at the units!\n\n200N/2500N/m = .08 m\n\n.08m is 8cm.
901904
I think here is a mistake:\n\n(1500 * 36) = 1000v\nv = 55.8 m/s\n\nyou have to put 500 instead 1500, because this is what we have after 1000 left us!\n\n(500*36)=1000v\nv=15 m/s
731116
Spring constant: k = 4π²m/T² = 4π²(.712)/(2*.685)² = 14.976 N/m\n\nω = √(k/m) = 4.58625 rad/sec\n\nVmax = Aω = .102*4.58625 = .4678 m/s
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units
965191
This isn't very hard - you just need to do each of the little pieces. I'll give some hints.\n1) When the block is at the extreme from the rest position all the system energy is stored in the spring. The stored energy is spring force (k*x) times spring deflection distance (x). So it's k*x^2 where x is 8.5. \n\nThe block has maximum speed when the block is in the center of the oscillation range (the spring is fully relaxed at that point), so all the energy is in kinetic energy. Energy is conserved, so\nMax spring energy = Max KE\nk*x^2 = (1/2)*m*v^2\n\n2) Best hint here: think, no math req'd.\n\n3) Spring force F = k*x. Acceleration = F/m\n\n4) Acceleration is max (see #3) at x = 8.5 and zero at x = 0. In the formula\nx = (1/2)*a*t^2,\nyou can use the average of the accelerations those 2 points for a.
858156
The braking force was enough to dissipate the kinetic energy.\n\nForce x distance = work = kinetic energy\n\nma x 160 = mv^2/2 , notice that m cancels out\n\n160a = v^2/2, solve for v...
905588
When the seven people get into the car it lowers to a new equilibrium position where the restoring force of the springs balances against the weight. Let m be the average mass of each individual then\n\nk*x = 7*m*g\n\nwhere k is the spring constant, x is the displacement caused by the seven people in the car, and g is the acceleration due to gravity.\n\nNow you can solve for the spring constant associated with the car\n\nk = 7*m*g/x\n\nThe bouncing car exhibits harmonic motion and is characterized by an angular frequency\n\nw = sqrt(k/M)\n\nwhere M is the mass on the unloaded car. Plugging in the known value for the spring constant\n\nw = sqrt(7*m*g/(M*x))\n\nThe linear frequency is\n\nf = w/(2*pi)\n\nwhich will be in Hertz.
832683
The energy required to stop the block is it's kinetic energy\n\nE = .5mv^2\n\nThe Energy stored in the spring is\n\nE = .5kd^2\n\nThe two are equivalent, so:\n\n.5mv^2 = .5kd^2\nmv^2 = kd^2\nv^2 = kd^2/m\nv = sqrt(kd^2/m) \nv= sqrt(750(0.012)^2/0.4) = .27 m/sec\n.
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
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
861868
what's the 30 g for? here's my answer:\n\nfrom F=m*a,\n\nF= (70 kg)*(3m/s^2)\nF=210 newtons...\n\nthis is the force needed to help the 70 kg person survive the crash...does that help?
378184
When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.\neverything in usual notation,\nacceleration A=R*a (a=angular acceleration)\nNewton's second law implies'\nMA=F-f         (resultant force)\nalso for torques,\nIa=FR+fR         (for the axis passing through the centre)\nFRom the two equatins,\n2F=mA+Ia/R\ni.e.  2F=maR+Ia/r\n   a=2F/(mR+I/R)\nI=0.5MR^2=0.05\nplugging hte values,\na=16 \ntherefore, A=aR\n            =16*0.1\n           A=1.6
309325
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)
658774
I would think you would use the force created by compressing the spring.\n\nThis force is given by the equation:\n\nF = k(Li - Lc)\n\nWhere:\nF = force\nk = spring constant\nLi = uncompressed spring length\nLc = compressed spring length\n\nThe below link is for a calculator for this force.\n\nhttp://www.efunda.com/DesignStandards/springs/calc_comp_Fstress.cfm
254925
I'm assuming that the pivot point will land between the 15 and 30 Kg loads because it seems more logical. If you try putting the X distance between the 20 and 15 loads, you'd probably get a negative distance. \n\nActually, the weights have to be multiplied by the pull of gravity (9.81m/ss) to make them actual forces but then again, they would all just cancel out.\n\ni hope you understand the crude diagram\n\n|-------10------|       - distance where weight of beam is applied\n|---------x-------|    - pivot point dist from left\n ____________________  - beam (20 m)\n|_________|__________|\n20_______15_________30 - applied forces (weights)\n\n20kg(X) + 15kg(X-10m)  = 30kg(20m-X)\n20kgX + 15kgX - 150kgm = 600kgm - 30kgX\n20kgX + 15kgX + 30kgX  = 600kgm + 150kgm\n                65kgX  = 750kgm\n                    X  = 750kgm/65kg\n                    X  = 11.53846 m from left end
564626
Which depends on the stres acted on the load and material used