When pushing off of a pool wall, a swimmer exerts a force parallel to the length
ID: 1605789 • Letter: W
Question
When pushing off of a pool wall, a swimmer exerts a force parallel to the length of her femur, compressing it 3 105 m. The bone is equivalent to a uniform cylinder 36.0-cm long and 1.75 cm in radius. Young's modulus for bone is 16 109 N/m2.
(a) Calculate the force exerted. __N
(b) If her mass is 72.0 kg and water resistance is negligible, what is her acceleration? Assume her weight is precisely supported by the water. __m/s2
(c) How fast is she going once she exerts this force through a distance of 20.5 cm, starting from rest? __m/s
Explanation / Answer
Given that Compression is e = 3*10^-5 m
Young's modulus is Y = 16*10^9 N/m^2
length of the bone is l = 36 cm = 0.36 m
radius is r = 1.75 cm = 0.0175 m
a) Using Y = stress / strain
Y = (F/A)/(e/l)
F/A = Y*(e/l) = 16*10^9*(3*10^-5/0.36) = 1.34*10^6
Force is F = 1.34*10^6*pi*r^2 = 1.34*10^6*3.142*(0.0175^2)
F = 1289.4 N
b) Fnet = m*a
accelaration is a = Fnet/m = (1289.4)/72 = 17.9 m/s^2
C) distance travelled is d = 20.5 cm = 20.5*10^-2 m
initial speed is Vo = 0 m/sec
a = 17.9 m/s^2
then using v^2 -Vo^2 = 2*a*d
V^2 - 0^2 = 2*17.9*0.205
V = 2.709 m/sec
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.