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eilaf · May 29, 2012, 11:47 PM

An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest?

ebaines · May 30, 2012, 06:22 AM

Originally Posted by eilaf

An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest?

From conservation of energy: as the man drops from height h=3 his velocity when he hits the water can be found from $\frac 1 2 mv^2 = mgh$ . Then when he hits the water you can apply Newton's 3rd law to determine force:

$<br /> F = m \frac {\Delta v}{\Delta t}<br />$

This assumes that the acceleration and force are constant (which may not be terribly realistic, but typically assumed for problems like this).

aishsure29 · Jun 11, 2012, 12:02 AM

A body started rolling over a horizontal surface with an initial velocity of 0.5 m/s.Due to friction,its velocity decreases at the rate of 0.05 m/s2. How much time it take for the body to stop?

ebaines · Jun 11, 2012, 05:36 AM

The definition of acceleration is change in velocity dividfed by time. Here you have the acceleration and change in velocity, so you can calculate the time required to make that change in velocity.