Answer to Question #128473 in Mechanics | Relativity for Osandi

Question

1A 50 kg gorilla is sitting on the limb of a tree 3 meters above the ground.
a) Find the potential energy of the gorilla sitting in the tree.
b) The gorilla jumps down from the tree limb to the ground. Use the conservation of energy to find the velocity of the gorilla just before hitting the ground.
c) Use the conservation of energy to explain why the total energy of the gorilla is 1470 J at 1 m above the ground.
d) The gorilla, feeling exuberant, throws a 0.50 kg banana straight up in the air from the ground with an initial energy of 30 J. How high will the banana travel up?

Answer 1

Solution.

$m=50kg;$

$h=3m;$

$g=9.8N/kg;$

$W_{kb}=30J;$

$a) W_p=mgh;$

$W_p=50kg\sdot9.8N/kg\sdot3m=1470J=1.47kJ;$

$b)W_p=W_k;$

$mgh=\dfrac{mv^2}{2}; gh=\dfrac{v^2}{2}; v=\sqrt{2gh};$

$v=\sqrt{2\sdot9.8N/kg\sdot3m}=7.67m/s;$

$c)$ According to the law of conservation of mechanical energy, it is known that the total mechanical energy in a closed system of bodies remains unchanged. Since, at a height of 3 meters, the gorilla had full energy $W_p=1470J$, it is the same at a height of 1 meter.

$d)W_{kb}=mgh_b\implies h_b=\dfrac{W_{kb}}{mg};$

$h_b=\dfrac{30J}{0.5kg\sdot9.8N/kg}=6.12m;$

Answer: $a)W_p=1.47kJ;$

$b)v=7.67m/s;$

$c)$According to the law of conservation of mechanical energy, it is known that the total mechanical energy in a closed system of bodies remains unchanged. Since, at a height of 3 meters, the gorilla had full energy W_p=1470J

Wp​=1470J, it is the same at a height of 1 meter.

$d)h_b=6.12m.$