a little spider monkey (let's call her flo) has stolen a bunch of bananas from her older brother charley and is now running away from him with the bananas in hand. she sees salvation in a long vine and grabs the bottom, hoping to swing away.
a) is flo, who has a mass of 5.3 kg, and the bananas, with a mass of 0.5 kg, reaches a height of 1.2 m before slowing down to a stop and swinging back down, how fast was she running when she grabbed the vine?
b) as she swings back down, charley (whose mass is 5.9 kg) is standing just at the bottom of the swing. he grabs her and they swing up together. what height will the two monkeys and the coveted bananas reach?
c) as a result of their fighting, they happen to drop the bananas just as the vine pendulum reaches this new height. how fast will the two monkeys be moving when they let go of the vine at the bottom again?
I got 4.8 m/s for the first answer. and 0.59 m for the second answer. Can someone please confirm these answers for me and also help me with the last question?
I haven't learned momentum
can you please explain what you meant?
Sure! I'd be happy to help explain the concepts involved and confirm your answers.
a) To determine how fast Flo was running when she grabbed the vine, we can use the conservation of energy equation. The initial potential energy when Flo reaches the highest point will be equal to the sum of her kinetic energy and the potential energy of the system at that point.
The initial potential energy is given by m₁gh₁, where m₁ is the total mass of Flo and the bananas (5.8 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h₁ is the height reached (1.2 m).
The kinetic energy is given by (1/2)m₁v², where m₁ is the total mass again and v is the velocity at that height.
Setting the initial potential energy equal to the sum of the kinetic energy and potential energy, we can solve for v:
m₁gh₁ = (1/2)m₁v²
(5.8 kg)(9.8 m/s²)(1.2 m) = (1/2)(5.8 kg)v²
67.68 joules = 2.9v²
v² ≈ 23.31 m²/s²
v ≈ √23.31 m/s
v ≈ 4.83 m/s (rounded to two decimal places)
Your answer of 4.8 m/s is correct!
b) To determine the height reached when Charley and Flo swing together, we can apply the principle of conservation of mechanical energy. At the lowest point of the swing, all the initial gravitational potential energy will be converted into the final gravitational potential energy and the kinetic energy of the system.
The initial potential energy when they swing back down is the same as the potential energy at the highest point reached previously, which is given by m₁gh₁ (where m₁ = 5.8 kg, g = 9.8 m/s², and h₁ = 1.2 m).
The final potential energy is given by the sum of the gravitational potential energy of the monkeys and the bananas, which is (m₁ + m₂)gh₂ (where m₁ = 5.8 kg, m₂ = 5.9 kg, g = 9.8 m/s², and h₂ is the height reached in this case).
Setting the initial potential energy equal to the final potential energy, we can solve for h₂:
m₁gh₁ = (m₁ + m₂)gh₂
(5.8 kg)(9.8 m/s²)(1.2 m) = (5.8 kg + 5.9 kg)(9.8 m/s²)h₂
67.68 joules = (11.7 kg)(9.8 m/s²)h₂
67.68 joules = 114.66 h₂
h₂ ≈ 0.59 m (rounded to two decimal places)
Your answer of 0.59 m is correct!
c) Finally, to determine the speed at which the monkeys let go of the vine, we can consider the conservation of mechanical energy again. At the lowest point of the swing, all the initial potential energy will be converted into kinetic energy.
The initial potential energy when the monkeys let go is the same as the potential energy at the height reached after the bananas were dropped, which is given by (m₁ + m₂)gh₃ (where m₁ = 5.8 kg, m₂ = 5.9 kg, g = 9.8 m/s², and h₃ = 0.59 m).
Setting the initial potential energy equal to the final kinetic energy, we can solve for the velocity:
(m₁ + m₂)gh₃ = (1/2)(m₁ + m₂)v²
(11.7 kg)(9.8 m/s²)(0.59 m) = (1/2)(11.7 kg)v²
63.8786 joules = 5.85v²
v² ≈ 10.94 m²/s²
v ≈ √10.94 m/s
v ≈ 3.31 m/s (rounded to two decimal places)
So, the approximate speed at which the monkeys let go of the vine is 3.31 m/s.
Well done on your answers! I hope the explanations helped clarify the concepts involved. Let me know if you have any further questions.