Calculating Forces in a Pulling System: Fp, Wp, Wm

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The discussion focuses on calculating the pulling force (Fp) in a system involving gravity (Fg) and friction (Ff). The equation Fp = Fg + Ff is questioned, particularly whether the force of friction should be subtracted from the force of gravity. The normal force (N) is identified as critical for determining the friction force, which is not simply equal to the block's weight. Participants are encouraged to decompose the weight of the block into components parallel and perpendicular to the incline to accurately calculate the normal force. Understanding these relationships is essential for correctly determining the forces in the pulling system.
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Homework Statement
A physics student pulls a block of mass m = 21 kg up an incline at a slow constant velocity for a distance of d = 4.5 m. The incline makes an angle Δ = 31° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.2. What is the work Wm done by the student?
Wm =
Relevant Equations
W = F*d
K = 1/2 mv^2
U = mgy
E = U + K
Fp = Fg +Ff
Fp = (21*9.81*sin(31)) + (0.2*21*9.81)
Fp = 147.30
Fp*d = 147.3 *4.5
Wp = Wm = 662.87 = 663 J

I am not sure what I am doing wrong. Should I subtract the force of friction from the force of gravity? Have I assumed that the pulling force is equal to the sum of gravity and friction when I shouldn't have? Should I use cos instead of sin?
 
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The friction force ##F_f## is equal to ##0.2N## where ##N## is the normal force from the incline to the block. What is the correct value of ##N##?? It is not 21*9.81 , because that's the full weight of the block...
Hint: Decompose weight of the block, into a component parallel to the incline and a component perpendicular to the incline. How does the normal force relate to the component perpendicular to the incline?
 
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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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