
#1
Nov407, 11:26 PM

P: 23

1. The problem statement, all variables and given/known data
A high diver of mass 83.0 kg jumps off a board 10.0 m above the water. If his downward motion is stopped 1.90 s after he enters the water, what average upward force did the water exert on him? 2. Relevant equations [tex]\sum[/tex]Forces = ma X=X[tex]_{0}[/tex]+V[tex]_{i}[/tex]+1/2at 3. The attempt at a solution the total force exerted by the diver on the water is: [tex]\sum[/tex]Forces = 83kg*9.8 (gravitational acceleration) = 813.4N then with the given time.. i don't know how to solve for the upward force 



#2
Nov407, 11:41 PM

P: 218

Would be easier to equate Potential Energy (P.E. = mgh, turned into Kinetic Energy) with work done (W = F x d).




#3
Nov507, 12:19 AM

P: 23





#4
Nov507, 12:35 AM

P: 218

forces, action and reaction pairs
OK. How about working out velocity when diver hits water and average deceleration after that?




#5
Nov507, 03:10 AM

P: 23

well, first I used: Xf = X0 + Vit + 1/2at
let Xf = 0, X0 = 10m , Vi=0 , a= 9.8m/s^2 therefore 0 = 10 + 1/2(9.8)t solve for t = 2.04 if Vf = Vi + at, since Vi=0, Vf = the velocity when it hits the water = (9.8)(2.04) = 19.992 m/s  for the second part (when the person enters water)  Vi=19.992m/s, Vf = 0, t=1.90 therefore, using the equation : Vf = Vi + at 0 = 19.992 + a(1.90) therefore a = 10.552 m/s^2 finally, force = mass * acelleration ==> force = 10.552*83kg = 875.816N Since I am suppose to convert the force into kN, do I do the following operation? 875.816N * (1kN/1000N) = 0.875816kN 



#6
Nov507, 04:47 AM

P: 15

Hi,
Your method, including the kN portion, looks correct (Assuming you didn't make any computation errors). However, you may want to use another formula (also derived from these two) which would allow you to skip a step. (Vf)^2 = (Vi)^2 + 2aX This would save you having to compute time. 



#7
Nov507, 07:20 AM

P: 218

Your answer should be no more precise than the data given. The question includes 83.0, 10.0 and 1.90. The first two indicate 3 significant figures; the last may indicate 2 or 3 so it's reasonable to take it as 3 for consistency.
So, assuming your answer is numerically correct, it should be given as 0.876 kN. Regards the distancevelocitiesaccelerationtime equations, it is helpful to know 5 forms, one omitting each variable. Using your variable symbols: No [itex] x [/itex]: [itex] v_{f} = v_{i} + at [/itex] No [itex] v_{i} [/itex]: [itex] x = v_{f}t  ½at^{\:2} [/itex] No [itex] v_{f} [/itex]: [itex] x = v_{i}t + ½at^{\:2} [/itex] No [itex] a [/itex]: [itex] x = ½(v_{i} + v_{f})t [/itex] No [itex] t [/itex]: [itex] v_{f}^{\:2} = v_{i}^{\:2} + 2ax [/itex] Usually you will know 3 of the five variables and be required to find one of the others; the easiest equation to use is the one that omits the 5th. For example, in this problem, when you want to find the divers' velocity on hitting the water water you know [itex] x [/itex], [itex] v_{i} [/itex], [itex] a [/itex] and [itex] x [/itex]. You want to find [itex] v_{f} [/itex]. So the easiest equation to use is the one that omits [itex] t [/itex]: [itex] v_{f}^{\:2} = v_{i}^{\:2} + 2ax [/itex] 


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