If a 58-kg person jumps out of a tree from a height of 2.1 m, calculate the force in the knee joints if?
157125
the force is mass * acceleration due to gravity= 58 x 9.8=568.4 newton\nbut when he lands the force is impact in nature i.e. knee will experience the force as a jerk which come to it in minimal small time, also impact force is twice of normal force. \nso force on knee = 568.2 x 2 =1126.8 newton.\nbut as per question knee has some stiffness so final force it will take is 1126.8/stiffness = answer
935163
Start with the formula, F = ma.  The force, F,  he is able to exert is equal to his weight times the coefficient of friction.  And his weight is his mass times the acceleration of gravity.  So, F = 0.9mg, and 0.9mg = ma.  Mass is on both sides, so it cancels.  0.9g = a, 0.9*9.81 m/s^2 = 8.83 m/s^2.  His final velocity, Vf = SQRT(Vi^2 + 2ad), where Vi is his initial velocity (zero), a is his acceleration, and d is the distance he travels.  So, Vf = SQRT(2*8.83*50) = 29.71 m/s.  Because his acceleration is constant, his average velocity is 29.71/2 = 14.86 m/s.  The time he takes is then d/Vave = 50/14.86 = 3.365 s.
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!!
982521
Well, when the ropes are horizontal, the child will be 2 m farther from the ground than when the ropes are vertical. (Draw a sketch if that will help.) \n\nGravitiational potential energy is m*g*h.\nYou already know that m*g is 400 N, and h is 2 m, so the increase in potential energy is 800 J.
40451
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)
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
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
1066612
Work = force x distance\n\nforce = mass x force of gravity\n\nso F=110 kg x 9.8 m/s = 1078 kg m/s^2 or 1078 Newtons\n\nWork = 1078 N x 12 m = 12936 kg m^2/s^2 or 12936 Joules
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1
894321
The horizontal component of speed\n= 40 m/s * cos(30)\n=40 cos(30) m/s\nmass\n=0.15 kg\nSince at the highest point, vertical velocity=0, and horizontal velocity is unchanged (ignore air resistance), so \nK.E.=mv^2/2=0.15*(40cos(30))^2/2\n=0.15*1600*3/4/2\n=90 joules
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
666057
*\n1/2 base X altitude + area (triangle)\n\n\n45 square centimeters
954796
51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
758260
I think, therefore i am not certain.....\n\nI know that F=ma, or W=mg. Therefore 18kg x really 9.81, but i'll use 10m/seconds squared. W = 180N, so normal force pushing UP on the lawn mover is 180N. If there was only a horizontal force it would take mew or 0.15 x normal fore so, 180N x 0.15. BUT there is a n angle so 35 is feta, 180N is the adjacent side. use trig function (cos) to find hypotenuse. cos35= 180N/ x. when  you get x, it equals the applied force to get the mover to keep on moving or sliding as "they" say. \n\n1.4 m/s will be present when force is applied!!!!!!!!!!
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
64624
less than that if the knee is in a locked stationary position.
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a)\n\nby dropping 3.2 meters tarzan lost \n\nPE=mgh=688x3.2= 2201.6 J  ,\n\nthis is converted to KE\n\n1/2 * m * V^2 , m=688/9.81= 70.13251784\n\nv=7.92 m/s, \n\nnow we can find centri acceleration\n\na=v^2/r = (7.923635529^2)/18 = 3.488 m/s2\n\nF=70.13 * 3.488 = 244.6 N\n\nso total force is 244.61344+688 = 932.61344 N  \n\nthis is less than 950 N\n\nthe vine doesn't break\n\nb) already found it in part a, max force was \n\n932.61344 N