How Does a Conveyor Belt's Kinetic Energy Change with Gravels?

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SUMMARY

The discussion focuses on the kinetic energy changes of a conveyor belt receiving gravels dropped at a rate of x (kg/s). The force required to maintain the belt's velocity v is calculated as F = x * v. The power generated in this system is determined by P = x * v^2. Additionally, the rate of change in kinetic energy is expressed as 1/2 * x * v^2, highlighting that this rate differs from power, particularly in the context of inelastic collisions.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with concepts of force and power in physics
  • Knowledge of kinetic energy and its calculations
  • Basic principles of inelastic collisions
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  • Learn about the conservation of momentum in collision scenarios
  • Explore advanced topics in dynamics related to conveyor systems
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Homework Statement


At a certain factory, gravels are dropped vertically from a machine at a rate of x (kg/s) onto a conveyor belt moving at velocity v.

a. what is the force needed to keep the belt moving at velocity v?

b. what is the power?

c. what is the rate of change in kinetic energy?

d. what is the difference between power and kinetic energy?

Homework Equations



F = dp/dt
P = F (dot product) v


The Attempt at a Solution



I got a, b, c,

a. F = dp/dt = dm/dt * v = x * v

b. P = F (dot product) v = x*v^2

c. rate of change in kinetic energy = 1/2 * dm/dt * v^2 = 1/2 *x * v^2

d. so rate of change in kinetic energy is not same as power!
what is going on?
 
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Supposed I told you that dropping gravel onto a conveyor belt is an inelastic collision?
 

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