Calculate the force applied and constant frictional force opposing its motion

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To calculate the force applied to the skateboard and the constant frictional force, Newton's second law (F(net) = m*a) is essential. The skateboard, weighing 0.50 kg, travels 1.0 m in 8.5 seconds, requiring the calculation of acceleration despite starting and ending at rest. The confusion arises from determining acceleration since the skateboard begins and ends its motion at rest. The discussion emphasizes the need to analyze the motion during the applied force and the subsequent coasting phase. Ultimately, applying kinematic equations will help resolve the acceleration and forces involved.
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A 0.50kg skateboard is at rest on a rough, level floor on which two lines have been drawn 1.0m apart. A constant horizontal force is applied to the shateboard at the beginning of the interval, and is removed at the end. The skateboard takes 8.5s to travel the 1.0m distance, and it then coasts for another 1.25m before coming to rest. calculate the force applied to the skateboard, and also the constant frictional force opposing its motion.
 
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What equations come into play here and what is your attempt at a solution?
 
Well I was thinking of the the Newton's second law, which is F(net)=m*a. Then I need acceleration. Which is confusing because both the velocities are at rest since it starts from a rest and ends at a rest. So I really don't know what to do!
 
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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