Box Acceleration and Velocity: Solving for Vertical Motion

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The problem involves a 5.0 kg box being pulled upwards with a force of 68 N. The acceleration of the box is calculated using the formula a = T/m - g, resulting in an acceleration of 3.8 m/s². After 2 seconds, the vertical velocity of the box is determined to be 7.6 m/s using the equation v = vo + axt. The calculations are confirmed to be correct. This demonstrates the application of Newton's second law and kinematic equations in solving vertical motion problems.
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Homework Statement


A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.

Homework Equations


ma = Force and a = T/m -g
a = 68/5 - 9.8 = 3.8 m/s^2

v = vo + axt
v = 0 + 3.8 x 2 = 7.6 m/s.

Does this look right?

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Yes, it does.
 
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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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