A person in a wheelchair approaches a 0.70-m-high ramp at a speed of 4.1 m/s. Calculate the final speed?
153350
This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
897942
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
290071
This was the best question I have seen on this site.  (That isn't saying much, so let me say it this way:  This is an excellent question!)\n\nThe answer is 100 steps. (Not considering the hidden steps since it's constantly revolving).\n\nFirst we have to consider the ACTUAL speed of each man since the escalator is moving, too.\n\n(Fast Speed) + (Escalator Speed) = (Fast Adj. Speed)\n(Slow Speed) + (Escalator Speed) = (Slow Adj. Speed)\n\nAnalyzing that if the fast man had been able to climb the entire time that the slow man did, he would have gone 150 steps, we can see that he finished in half the time it took the slow man (because 75/150 = 1/2).  Therefore we also have:\n\n2 (Fast Time) = (Slow Time)\n\nWe also use the simple equation defining speed (s=d/t) to determine that:\n\n(Fast Adjusted Speed) = (Distance) / (Fast Time)\n(Slow Adjusted Speed) = (Distance) / (Slow Time)\n\nwhere (Distance) is equal to the actual distance traveled (eg. the distance from the top of the escalator to the bottom of the escalator).\n\nInserting the few givens and abbreviating the variables, we have, respectively:\n\nEq 1) 3 + ES = FAS\nEq 2) 1 + ES = SAS\nEq 3) 2 FT = ST\nEq 4) FAS = D/FT\nEq 5) SAS = D/ST\n\nThis looks like 6 variables with only 5 equations, but it is still actually solvable.\nUsing Eq.3 to simplify Eq.4 and Eq.5:\n\nEq 4) FAS = D/FT\nEq 5) SAS = D/2FT\n\nand we can reduce our number of variables by simply saying that X = D/FT, hence,\n\nEq 4) FAS = X\nEq 5) SAS = X/2\n\nSo now we use these to analyze Eq.1 and Eq.2:\n\nEq 1) 3 + E = X\nEq 2) 1 + E = X/2\n\nTwo variables, two equations.  E turns out to be 1.  This means that the escalator moves at the same speed as the slow man (remember that we assigned his abitrary, but proportional speed to be 1 in the equations).  If he and the escalator are both the same speed, and he contributes 50 steps to the journey, then the escalator must also have contributed 50 steps, totalling 100 steps.\n\nOr guess and check works pretty well, too.
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
985855
velocity = df(t)/dt = 4π + 4πcos(4πt)\nacceleration = dv(t)/dt = -16π^2sin(4πt)\n\nParticle is at rest when velocity = 0, or 4π+4πcos(4πt) = 0\n1+cos(4πt) = 0\ncos(4πt) = -1   The values for which cosø = -1 are ø=n(odd)π, so when\n4πt=n(odd)*π, t = 1/4*n(odd); thus for t = 1/4, 3/4, 5/4, 7/4, 9/4, 11/4 the velocity is zero\n\nThe maximum velocity occurs when cosø = 1, so the max velocity is 8π
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
418402
Using one of the equations of linear motion:\n\nv² = u² + 2as\n\nInitial speed u = 0\nAcceleration a = 32 ft/s\nDistance s = 9 ft\n\nSolve for final velocity v = sqrt( 2 * 32 * 9) = 6.2 ft/s\n\nThis assumes negligible air friction.
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
1053261
70cm hieght of the topstep\n4 degree angle off the ground\n\na) 70/(sin 4*) = 1003 cm or 10.03 m\n\nb) arcsin times 70/900 = 4.46* \n\nyes it will meet wheel chair code
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
1151145
Calculate acceleration a=(24-10)/16 = 14/16\nF=ma=1200*(14/16)=1050N
750320
If it takes 6 secs to reach the max, it will take an additional 6 secs to reach your starting point.  In this case it is 2560.  Then it is a matter of finding the time from that point to going to the ground.\n1st lets correct the orginal equations (acceleration due to gravity is down/negative)\nh = -16t^2 + 192t + 2560\n2nd - set the equation to zero since we want to find the time to ground.\n0 = -16t^2 + 192t + 2560\n16t^2 - 192t = 2560\nt^2 - 12t = 160\n(t^2 - 12t + 36) - 36 = 160\n(t - 6)^2 = 196\nt - 6 = 14\nt = 20\nConclusion: 6 seconds to reach max ht of 3136, another 6 seconds to reach starting point of 2560 and another 8 seconds to hit the ground.
755607
To solve this problem, you need to use conservation of energy.  At the beginning, the block is at rest at the top of this ramp.  At the bottom of the ramp, the block still has some kinetic energy, but some of it has dissipated due to friction.  Thus, the equation that governs this problem is:\n\nInitial Potential Energy = Dissipated Energy due to Friction - Final Kinetic Energy\n\nSince Potential Energy is defined only in the vertical direction, we must use the vertical components of the energy.\n\nWe can find the height of the ramp using trigonometry:\n\nsin(theta) = height/incline -----------> height = incline*sin(theta)\n\nheight = 3.03*sin(28.8) = 1.46 m\nPotential Energy = mgh = (2.51)(9.81)(1.46) = 35.95 J\nFrictional Energy (vertical) = u*m*g*sin(theta)*h\nKinetic Energy (vertical) = 0.5*m*[v*sin(theta)]^2 = (0.5)(2.51)(1.77sin(28.8))^2\n\nKinetic Energy = 0.91 J\n\nPlugging these values into the equation, you get:\n\n35.95 = u*m*g*sin(theta)*h + 0.91\n\nWe have all the values, now solve for u\n\n---------------\n\nHope this helps
884244
a. 1.10 MJ\n\nEnergy = mass * acceleration * distance = m * a * s\n\nm = 1600 kg\nAssume that the car is accelerating at 1 m/s^2 =>  a = 1\n\nIt would take 25 seconds for a car accelerating at 1 m/s^2 to go from 15 m/s to 40 m/s.\n\nThe average velocity during the acceleration is (15 + 40)/2 = 27.5 m/s.\n\nThe distance, s, is 27.5 m/s * 25 seconds = 687.5 m\n\nm*s*a = 1600 kg * 1 m/s^2 * 687.5 m = 1.1 e6 J = 1.1 MJ\n\nIf you assume another acceleration (let's say 5 m/s^2) you will get the same because the same amount of work is being done.