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Force exerted to stop falling object 
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#1
Oct2004, 09:54 PM

P: 10

Having trouble with this problem:
3. [SFHS99 5.P.37.] A 50.0 kg diver steps off a 12.0 m high diving board and drops straight down into the water. If the diver comes to rest 5.0 m below the surface of the water, determine the average resistance force exerted on the diver by the water. Seems to me like a simple kinematic problem, so I used kinematic equations to find the Vf when it hits the water (15.32m/s), then used that as a new Vi to find the time it took to decelerate from Vi to 0 in 5 meters. (.653 sec) I then used that time to find a (de)acceleration rate for the diver, for which I got ()23.46 m/s^2. I then simple multiplied that by the diver's weight to get a resistence force, which was ()1173 N. However, my online homework is telling me this is wrong! Any tips? 


#2
Oct2004, 10:44 PM

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P: 5,197

You know her change in kinetic energy. (0  1/2mvf^2) It is equal to the work done (negative in this case) on her body from the surface of the water to 5.0m below. If I tell you the work done on an object and the distance through which it was done, what can you tell me about the force acting on the object across that displacement? Assuming the force is constant, or taking it to be the average force?
With the method you used...how did you calculate the time it took to slow to a stop underwater if you didn't know the acceleration in the first place? 


#3
Oct2104, 05:56 AM

P: 10

ok, so I used 1/2mVf^2 to find the work done on the divers body, and got 5867.6 J. Then, since W = Fd, I divided the work done by the distance over which it was done, and got 1173.5 N. Tried both positive and negative of that, but online homework is still telling me its wrong!



#4
Oct2104, 05:59 AM

P: 10

Force exerted to stop falling object
Btw to find the time it took to slow to a stop, i used the kinematic equation Xf = Xi + 1/2(Vf+Vi)t



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