Solving for Tension in an Elevator with Acceleration of 3.0 m/s^2 | Physics Help

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SUMMARY

The discussion focuses on calculating the tension in a cable supporting an elevator weighing 25,000 N while it accelerates upward at 3.0 m/s². The key formula involves understanding that the net force acting on the elevator is the sum of gravitational force and tension. Given that gravitational acceleration (g) is 9.8 m/s², the tension must exceed the gravitational force to account for the upward acceleration. The net force equation, ΣF = m*a, is essential for solving this problem.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Knowledge of gravitational force calculations
  • Ability to convert weight (N) to mass (kg)
  • Familiarity with basic kinematic equations
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  • Learn how to calculate mass from weight using the formula: mass = weight/g
  • Study the concept of net force and its application in various physical scenarios
  • Explore examples of tension calculations in different acceleration contexts
  • Review the principles of forces acting on objects in motion
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This discussion is beneficial for physics students, educators, and anyone seeking to understand the dynamics of forces in motion, particularly in elevator systems.

Aiasha
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guys iam takin physics
its a takhom test
nd i don't get physics attt allll!
pllz hlp me with this
i need to show work as well

an elvtor weighing 25,000 N is supported by a steal cable. that is the tension in the cable wehn the elavotor is begin accelerated upward at the reate of 3.0 m/s^2? (g= 9.8m/s^2)
pleasezz help[
 
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TIP: The tension in the cable must be greater than the gravity pulling down. Think about that.

Also remember that if this is an isolated system where only tension and gravity are acting on the elevator, the net force is equal to the gravity + tension. Is there a way you can find the mass? Once you find the mass, might there be also be a way to calculate the netforce using mass and the acceleration? REmember, Netforce=Gravity+Tension
 
remember

[tex]\sum \overline{F} = m\overline{a}[/tex]

directions are imp.
 

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