Recent content by jphillip

Homework Statement The Femur bone in the human leg has a minimum effective cross section of about 3.0 cm2 How much compressible force can it withstand before breaking Homework Equations Compressive Strength (bone)(N/m2) = 170 x 106 P = F/A The Attempt at a Solution Not sure what a...

Homework Statement 65 kg trampoline artist jumps vertically upward from the top of a platform with the speed of 5.0 m/s. a.)How fast is he going as he lands on the trampoline and b.) If the trampoline behaves like a spring with stiffness of 6.2 X 104 N/m how far is it depressed Homework...

I have not made it to conservation of energy in Class. Thanks for the help I am sure I will be back for more. I am really enjoying this class, just not the summer version of I believe LOL.

Here are the formulas that I have used... initial height = 1.80m N - Newtons kg - kilograms (Block 1 31.31N - Block 2 21.56N) / (block 1 3.3kg + block 2 2.2kg) = a So (31.36N - 21.56N) / (3.2kg + 2.2kg) = 1.8148 m/s2 Then solve for Velocity using v2 = v2i + 2a(y) V2= 0 +...

\sqrt{2*1.81*1.80} = v = 2.55 2.552 / (2*9.8m/s2) = y = .332 + 3.6 (1.80(starting height)+1.80 accl height) total height 3.93m

So that gets me the velocity2 I would then have to solve for the additional distance correct

O.K. V2 = 2(1.81m/s2)(1.80m)

I believe F = ma which worked out a = F/m So 3.2kg - 2.2kg = 1kg * 9.8 = F (3.2 + 2.2) = m ? (not sure here) I think

I got the first forumla from an example in my book that appeared to be calculating what I thought I needed. An elevator question, what did I miss or am I missing

two blocks (2.2kg and 3.2kg) both hanging 1.80 m from the ground on a pulley(massless / frictionless) that is 4.80 m from the ground. What is the maximum height the smaller block (2.2kg) reaches, when the system is released? It appears that I need to find its velocity first of the system...