Calculating Forces and Power on a Block on an Inclined Plane

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The discussion focuses on calculating forces and power for a block on a frictionless inclined plane at a 25-degree angle with a mass of 2.6 kg. The force on the string is determined to be 10.4 N, as it equals the parallel component of the block's weight. When the block moves at a constant speed of 0.85 m/s, the force on the string remains the same due to balanced forces, as explained by Newton's First Law. The power required to maintain this speed is calculated to be 8.8 W, while the rate of change of gravitational potential energy (GPE) is found to be 9.15 W. The calculations confirm the correct application of trigonometric principles in determining the forces involved.
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A block is held stationary on a frictionless inclined plane by means of a string.

The angle of inclination of the plane is 25 degrees, the block has mass 2.6 kg. Calculate the force on the string, you may assume that g = 9.8 m/s2 . Sorry for the poor quality of the attachment but I think you get the idea.

So, its the first time I came across this type of question so I'm not sure whether what I did is correct or not:

a) So the force on the string: Block not moving, thus, equal to parallel component of the weight, hence:

T = sin 25 x (2.6 x 9.8)
T = 10.4 N

b) The string is pulled so that the block is now moving at a constant speed of 0.85 m/s up the inclined plane: Explain why the magnitude of the force on the string is the same now as it was in the first question?

N1 First Law says that if forces are balanced, object must be at rest or moving at a constant velocity, thus, since the weight didn't change, the force on the string must be the same

c) Calculate the power required to move the block at this speed:

P = F x velocity
P = (sin 25) x (9.8 x 2.6) x 0.85
P = 8.8 W

d) State the rate of change of GPE of the block:

So, in 1 second, since moving at 0.85 m/s, block is displaced 0.85 m in one second at an angle of 25. Thus, vertical displacement in one second = sin 25 x 0.85
hence: (mxgxh) = (2.6 x 9.8 x (sin 25 x 0.85)) = 9.15 W

Thanks in advance,
Peter G.
 

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Looks good to me.
 
Thanks, I was afraid that, when calculating the force on the string, thus, finding the parallel component of weight I had not used the right trig law, since that could alter all my other answers. So I did that right?
 
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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