Question

Abasket ball player ol mass q5 kg lands onthe court on one Foot aPter Jumping to a height啅0.44m. His Femur contracts 2.0 in comina to astop aj use conservahan of enensy to determine the spring constant b) calculata his aceleration as he comes to acomplete stop ouer a human Femur those一2.0-cm. E1- K 2. Kamav 2 amah ケ(95X2.ciitees)(93)(-YY 0.02) 2 294mls p) qu 2 4ye 2a(0.02) 2(0.0 2)06 m/s 2

Answer 1

A)

Conserving energy :

All the initial gravitational PE stored in the body is converted to spring PE stored in the femur

So, mgh = 0.5*k*x^2

here, x = compression = 2 cm = 0.02 m

h = 0.44 m

k = spring constant

So, 95*9.8*0.44 = 0.5*k*0.02^2

So, k = 2.05*10^6 N/m

NOTE : Your energy equation seems to be wrong, the initial gravitaitonal PE is first converted to KE and this KE is again converted to spring PE. So, all gravitational PE is indirectly converted to spring PE

B)

Using equation of motion:

v^2 = u^2 + 2as

So, 0^2 = 2.94^2 + 2*a*0.02

So, a= -2.94^2/(2*0.02) = -216.1 m/s2 <------- answer