What is the kinetic energy of object 1 after being pulled?

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The discussion focuses on the kinetic energy of two objects pulled by the same force over the same distance. Object 1 has a work done of 500J, resulting in a kinetic energy of 500J after being pulled. Object 2, experiencing the same force and distance, also has a work done of 500J, leading to the same kinetic energy. The ratio of the mass of object 1 to object 2 is determined to be 1:4, based on their velocities and kinetic energy equations. Understanding the underlying concepts simplifies the problem-solving process.
AdnamaLeigh
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I'm not sure if I have these concepts correct for lab...

Two objects of different mass start from rest, are pulled by the same magnitude net force, and are moved through the same distance. The work done on object 1 is 500J. After the force has pulled each object, object 1 moves twice as fast as object 2.

a) How much work is done on object 2?
W1 = F1D1 = 500J thus
W2 = F1D1 = 500J

b) What is the kinetic energy of object 1 after being pulled?
Wnet = ∆K = .5mv^2 – 0 = 500J

c) What is the kinetic energy of object 2 after being pulled?
Wnet = ∆K = .5mv^2 – 0 = 500J

d) What is the ratio of the mass of object 1 to the mass of object 2?
Ob1: ∆K = .5m(2v)^2 = 4(.5mv^2)
Ob2: ∆K = .5m(1)^2 so...
1:4

Did I do this correctly? It seemed too easy...
 
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Looks good to me. As far as it being easy, it only is if you understand the concepts with which you're dealing. Don't make problems any harder than they need to be.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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