Calculating Kinetic Friction on an Amusement Park Ride

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Homework Help Overview

The problem involves calculating the work done by kinetic friction on a seal sliding down a ramp into a pool. The ramp's height and angle are provided, along with the seal's speed upon reaching the water. The context is rooted in the principles of work and energy, specifically relating to frictional forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between work and energy, with attempts to apply conservation of energy principles. Questions arise regarding the calculation of work done by friction and the implications of mechanical energy not being conserved.

Discussion Status

There is an ongoing exploration of the concepts involved, with some participants providing guidance on the relationship between work, force, and distance. Multiple interpretations of the problem are being considered, particularly regarding the role of friction in energy loss.

Contextual Notes

Participants note that the mechanical energy appears not to be conserved, suggesting that the work done by friction is a key factor in understanding the problem. The angle of friction's effect on work is also under discussion.

map7s
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A 30.0 kg seal at an amusement park slides down a ramp into the pool below. The top of the ramp is 1.60 m higher than the surface of the water and the ramp is inclined at an angle of 30.0° above the horizontal. The seal reaches the water with a speed of 4.90 m/s. What is the work is done by kinetic friction?
I tried a number of things, but I wasn't sure how to calculate kinetic friction.
 
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What did you try? This problem is about work and energy. It just happens to be a frictional force doing some work in this problem. All you need to know is the work done by the force, and how that is related to the force and distances involved.
 
I tried doing W=Fdcos(x). I tried that after I did a conservation of energy equation mgh=1/2 mv^2 but I didn't know where to go with that.
 
map7s said:
I tried doing W=Fdcos(x). I tried that after I did a conservation of energy equation mgh=1/2 mv^2 but I didn't know where to go with that.
I assume that if you calculated mgh at the top of the slide, and 1/2mv^2 at the bottom you found that mechanical energy was not conserved in this problem. The difference is the work done by friction. Friction acts opposite the direction of motion, so the angle in the dot product is 180 degrees. The work done by friction is negative; it takes mechanical energy from the seal.
 

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