How Is Work Done by Friction Calculated in a Spring-Box System?

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SUMMARY

The work done by friction in a spring-box system can be calculated using the equation W = F · d, where F represents the friction force and d is the distance traveled. In this scenario, the spring constant is 200 N/m and the spring is compressed by 20 cm (0.2 m). The initial energy stored in the spring, given by KE_{spring} = ½kΔx², equates to the work done by friction when the box comes to a stop. Therefore, W_{friction} = 20 J, calculated as W_{friction} = ½ * 200 N/m * (0.2 m)².

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of kinetic energy equations
  • Familiarity with work-energy principles
  • Basic concepts of friction in physics
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  • Learn about the coefficient of friction and its impact on work done
  • Explore advanced applications of the work-energy theorem
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Homework Statement


A horizontal spring with spring constant 200 N/m is compressed 20 cm and used to launch a box across a rough horizontal surface. After traveling a distance it stops. What is the work done by friction force?

Homework Equations


W = F \cdot d
KE_{spring} = \frac{1}{2}k \Delta x^2

The Attempt at a Solution


I don't know where to start, because I'm not sure what other relevant equations I can use. I'm thinking that the friction has to do enough work to make the box stop (v = 0). But I don't know how to relate the work by friction to the energy of the spring.
 
Last edited:
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Initially you have the spring energy. It gets converted into kinetic energy. That gets converted into heat by the work of friction. All your initial energy is equal to the work of friction.
 
Ah... so W_{friction} = KE_{spring} = \frac{1}{2}k \Delta x^2 = \frac{1}{2} \cdot 200 \cdot .2^2?
 
Last edited:
looks good!
 
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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