Work Done on a Block on an Inclined Plane

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The discussion focuses on calculating the work done on a block of mass 18kg pushed up an inclined plane with a force of 150N, considering friction. The work done by the applied force is determined by its components along the incline, factoring in friction's effect on acceleration. The gravitational work is calculated using the sine of the incline angle, while the normal force does no work due to lack of displacement in its direction. The impact of friction is clarified as it slows the block's acceleration but does not change the fundamental work calculations. The final part of the discussion involves finding the block's velocity after moving 5m, using the net work done, which includes friction.
gunjay
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Homework Statement


A block of mass m=18kg is pushed horizontally with a force of Fp=150N up an inclined plane of angle θ=32° and coefficient of friction of μ=0.10, a distance of x=5m. a) What is the work done by Fp. b) Work done by the gravitational force. c) Work done by the normal force.


Homework Equations


W = ∫Fdl
Force in the same direction as displacement.
or W = F*x

The Attempt at a Solution


My main question is how is work affected by friction?
This question is like a part 2 where the first one had no friction and this one does but i don't see how friction gets involved at all.

a) The Fp has 2 components, one that moves parallel to the inclined plane and the other perpendicular to the plane. so Fpx=Fpcosθ. So including friction do i do W = (Fpx-Ff)*x or just W=Fpx*x or ?

b) Wg=∫Fgdx = mgsinθ*x

c) WN=∫FNdx = 0 = there is no displacement in the direction of the normal force.
 
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a) You got it with your second equation there. It's irrelevant whether you are fighting against gravity, friction, or little green men pushing back in the opposite direction. Work is work.

b) and c) You got it. Friction only causes the block to accelerate slower.
 
I believe i found the answer.
There was a part d that i thought didn't apply but it did. d) asks for the velocity of the block after it moves those 5m. Using KE= net W i was able to find an answer using friction that also seems plausible.
 
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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