Force of 1.5kg Crate Falling onto Spring Scale

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SUMMARY

A 1.5kg crate falling from a height of 2.0m onto an industrial spring scale with a spring constant of 1.5 x 10^5 N/m results in a potential energy of 30J. The force exerted by the crate upon impact is calculated as 14.75N. To analyze the system effectively, it is recommended to approach the problem using conservation of energy principles, first determining the kinetic energy (KE) just before impact and then the potential energy (PE) stored in the spring at maximum compression.

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  • Understanding of gravitational potential energy (PE) calculations
  • Knowledge of kinetic energy (KE) concepts
  • Familiarity with Hooke's Law and spring constants
  • Basic principles of conservation of energy
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blackout85
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A 1.5kg crate falls from a height of 2.0m onto an industrial spring scale with a spring constant of 1.5 X 10^5 N/m. At its greatest compression the reading on the scale is:

My work:

mgh= PE
(1.5kg*9.81m/s*2.0m)= 30J
The potential energy of the crate is 30J
The Force of the crate is 1.5kg* 9.81m= 14.75
Would the force of the crate be equal to the force of compression

Thank you
 
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I'm not sure what you mean by "force of the crate." I also don't think using forces is a good way to solve this problem. You did a good job finding the PE at the top.

There are two ways to go from here. The better way (and also the slow way) is to turn it into two conservation of energy problems: first find the KE it attains just before hitting the spring, then find the PE of the spring after it comes to a stop.
 

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