Question
Please show all work and calculations in very clear steps.
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Problem 4:
Kangaroos are very efficient at hopping; they can hop for miles without fatigue. Their legs have very long tendons that act like springs: they stretch and relax on each hop, storing and releasing elastic energy. A kangaroo might have a mass of m = 25 kg and two tendons (one in each leg) that are 30 cm long. The effective spring constant of a kangaroo tendon is about k ~ 4 × 105 N/m a) When fully stretched, the tendons increase in length by 2.4 cm. How much elastic energy is stored in two fully-stretched tendons? b) Consider a kangaroo at rest whose tendons are fully stretched (storing energy as described in part a If all of that elastic energy propels the kangaroo straight up, and the kangaroo's muscles do not do any work, how fast will the kangaroo be traveling when she initially leaves the ground? (You may assume that the work done by gravity during that push-off phase is negligible.) How high will she jump? c) Suppose that instead of trying to jump as high as possible, the kangaroo "hops" in place (bouncing up and down). If the kangaroo's hopping is purely based on the elasticity of tendons, what h frequency would you expect? Assume that the hopping motion can be approximated as simple harmonic motion (hint: do problem 1 first). d) It has been proposed that kangaroos use their leg muscles to modify the effective spring constant of their tendons: the actual hopping frequency of kangaroos is roughly 2 Hz. What should be the modified spring constant (of one tendon) in order for the harmonic motion of hopping to match the actual frequency of hopping?
Explanation / Answer
Since there are two tendons hence the effective spring constant of each tendon will be, k
k= 4*105 +4*105 =8*105 N/m
a) elastic energy stored in them,E= 0.5*k*x2
given x=0.024 m
E=0.5*8*105*0.0242 = 230.4 J
b) This elastic energy will first be converted to its kinetic energy and then its potential energy
E=KE=PE
E= 0.5*m*v2; E=mgh ;
using both equation we get,
v=sqrt(2*E/m)=sqrt(2*230.4/25) = 4.29 m/s
h=E/(mg) = 230.4/(25*9.8)= 0.940 m
c) Its angular frequency would be, w=sqrt(k/m)
w=sqrt((8*105)/25)=178.88 rad/sec
and w=2*3.14*f
f=w/(2*3.14)= 178.88/(2*3.14)= 28.48 Hz
d) given, f=2 Hz
w=2*3.14*w = 12.56
k'=mw2=25*12.562=3943.84 N/m
This is for two tendons
for one tendon it will be 3943.84/2
=1971.92 N/m
