Energy with nonconservative forces

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SUMMARY

The discussion focuses on calculating the speed of a block transitioning from a horizontal surface to an inclined ramp under the influence of nonconservative forces, specifically friction. The block, weighing 3.0 kg and initially moving at 7.0 m/s, slides 2.0 m before reaching a ramp inclined at 40 degrees with a coefficient of kinetic friction of 0.30. The approach involves applying the work-energy principle, where external work is zero, and analyzing changes in kinetic and thermal energy. The calculations for both the speed at the ramp and the distance slid on the incline are confirmed as valid by participants.

PREREQUISITES
  • Understanding of the work-energy principle in physics
  • Knowledge of kinetic friction and its effects on motion
  • Familiarity with trigonometric functions related to inclined planes
  • Ability to perform energy conservation calculations
NEXT STEPS
  • Study the effects of different coefficients of friction on motion
  • Learn about energy conservation in systems with nonconservative forces
  • Explore advanced applications of the work-energy theorem
  • Investigate the dynamics of objects on inclined planes with varying angles
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Students and educators in physics, engineers working with mechanical systems, and anyone interested in understanding the principles of energy conservation and motion involving friction.

maniacp08
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A 3.0kg slides along a horizontal surface with a speed of 7.0m/s.
After sliding a distance of 2.0m, the block makes a transition to a ramp inclined at an angle of 40 degrees to the horizontal. The coefficient of kinetic friction between the block and the surfaces is .30.

Find
a) the speed of the block when it reaches the ramp
b)the distance the block slides along the inclined surface before coming momentarilly at rest(neglect any energy dissipated along the transition curve)

For part A I used
External Work = change in potential energy + change in kinetic energy + change in thermal energy

External work = 0
change in potential energy = 0

so it becomes
0 = change in kinetic energy + change in thermal energy

change in kinetic energy = 1/2 m Vf^2 - 1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2
I plug in the numbers and I solve for Vf?

For Part B:
I can use the same equation but this time I am solving for height.
External Work = change in potential energy + change in kinetic energy + change in thermal energy

External Work = 0
change in potential energy = -mgh
change in kinetic energy = -1/2 m Vi^2
change in thermal energy = Uk * FN * displacement = Uk * m * g * 2

Solve for the height it reaches, then use trig to find the distance/hypotenuse.

Are these approaches correct? Thanks for helping.
 
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Yes, sure, the energy conservation is a good approach. Good luck!
 

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