Forces, action and reaction pairs

In summary: So, using the 4 known values, v_{f}^{\:2} = 0 + 2(-9.8)(10) and, since we want the square root of v_{f}^{\:2} we get v_{f} = √[0 - 2(-9.8)(10)] = √[196] = 14.00. The negative root doesn't make sense here.You can now use v_{f} and v_{i} to find the average velocity through the water. You can also use the average velocity and the time through the water to find the average acceleration through the water. And you can use that to
  • #1
mkwok
23
0

Homework Statement


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?


Homework Equations


[tex]\sum[/tex]Forces = ma
X=X[tex]_{0}[/tex]+V[tex]_{i}[/tex]+1/2at



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
 
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  • #2
Would be easier to equate Potential Energy (P.E. = mgh, turned into Kinetic Energy) with work done (W = F x d).
 
  • #3
catkin said:
Would be easier to equate Potential Energy (P.E. = mgh, turned into Kinetic Energy) with work done (W = F x d).

sorry, but I haven't learned that, and I'm not suppose to know how to do it that way
 
  • #4
OK. How about working out velocity when diver hits water and average deceleration after that?
 
  • #5
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
 
Last edited:
  • #6
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
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 distance-velocities-acceleration-time 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]
 

Related to Forces, action and reaction pairs

What is an action-reaction pair?

An action-reaction pair is a pair of forces that occur simultaneously and act in opposite directions on two different objects.

What is Newton's Third Law of Motion?

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when two objects interact, the forces they exert on each other are equal in magnitude and opposite in direction.

Can an object have an action-reaction pair acting on itself?

No, an object cannot have an action-reaction pair acting on itself. Action-reaction pairs always involve two different objects.

Do action-reaction pairs cancel each other out?

No, action-reaction pairs do not cancel each other out. While they have equal and opposite forces, they act on different objects and have different effects.

Can you give an example of an action-reaction pair in everyday life?

One example of an action-reaction pair in everyday life is the recoil of a gun. When a bullet is fired, the gun experiences a force in the opposite direction, causing it to recoil.

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