Question

A 3.7 kg bird traveling 7.99 m/s 50.3 m off the ground swoops down to 8.13 m off the ground. At this point it is traveling 23.2 m/s. What amount of energy is lost to air resistance?

Answer 1

Solution.

$m=3.7kg;$

$v_1=7.99m/s;$

$h_1=50.3m;$

$h_2=8.13m;$

$v_2=23.2m/s;$

$\Delta$ $W-?;$

$\Delta W=W1-W_2;$

$W_1=mgh_1+\dfrac{mv_1^2}{2}=m(gh_1+\dfrac{v_1^2}{2});$

$W_2=mgh_2+\dfrac{mv_2^2}{2}=m(gh_2+\dfrac{v_2^2}{2});$

$W_1=3.7kg\sdot(9.81m/s^2\sdot50.3m+\dfrac{(7.99m/s)^2}{2})=1943.8J;$

$W_2=3.7kg\sdot(9.81m/s^2\sdot8.13m+\dfrac{(23.2m/s)^2}{2})=1290.8J;$

$\Delta W=1943.8J-1290.8J=653J;$

Answer: $\Delta W=653J.$