Solving Work Negative Sign Homework Questions

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SUMMARY

The discussion focuses on solving homework questions related to work and energy in physics, specifically addressing the negative sign in the work done by friction. The equation -μmg(50) is used to represent the work done against friction, where μ is the coefficient of kinetic friction, m is mass, g is gravitational acceleration, and 50 ft is the displacement. The negative sign indicates energy loss due to friction as the cars decelerate. Additionally, an error in the calculation for car 2's stopping distance is noted, emphasizing the correct use of 20 ft instead of 50 ft.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of work and energy
  • Knowledge of kinetic friction and its coefficient
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of conservation of energy in physics
  • Learn about the coefficient of kinetic friction and its applications
  • Explore the derivation of work-energy principles
  • Practice solving problems involving deceleration and stopping distances
USEFUL FOR

Students studying physics, particularly those tackling work and energy problems, as well as educators looking for clarification on friction-related concepts in mechanics.

xzibition8612
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Homework Statement


See attachment


Homework Equations





The Attempt at a Solution



Why is it -umg(50)? Why negative? Also I think this has something to do with conservative force and friction, and I'm very confused on this. Any help would be appreciated thanks.
 

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xzibition8612 said:

Homework Statement


See attachment

Homework Equations


The Attempt at a Solution



Why is it -umg(50)? Why negative? Also I think this has something to do with conservative force and friction, and I'm very confused on this. Any help would be appreciated thanks.

When the force is constant, Work = Force*Displacement.

In this case, it's negative because energy is being taken away from the cars by the frictional force, while they're being decelerated to a stop. The energy dissipated by friction is equal to the drop in kinetic energy (as each car comes to rest). This is conservation of energy.

μ is the coefficient of kinetic friction. mg is the weight (equal to the normal force on the car). 50ft is the distance through which car 1 moves.

BTW, there's an error in the second part of the solution. The expression for car 2 should be \sqrt{(2)(20)(0.5)(32.2)} (i.e. 20, not 50), since 20ft is the distance car 2 takes to be brought to rest. The final stopping distance was calculated correctly, though.
 

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