Question

A spring with a 1.2x10^4 N/m force constant is stretched 6 cm from its equilibrium position. Calculate the elastic potential energy of this spring. PE = (1/2)kx^2 PE = (1/2)(1.2x10^4 N/m)(0.06 m)^2 PE = (1/2)(1.2x10^4 N/m)(0.0036 m^2) PE = 21.6 J

Name: a spring with a 12x104nm force constant is stretched 6cm from its equilibrium position calculate the elastic potential energy of this spring pe joules 27198
Uploaded: 2022-11-29T14:40:37-08:00
Duration: 1 min 10 s
Channel: Madhur L
Description: a spring with a 12x104nm force constant is stretched 6cm from its equilibrium position calculate the elastic potential energy of this spring pe joules 27198

          A spring with a 1.2x10^4 N/m force constant is stretched 6 cm from its equilibrium position. Calculate the elastic potential energy of this spring.
PE = (1/2)kx^2
PE = (1/2)(1.2x10^4 N/m)(0.06 m)^2
PE = (1/2)(1.2x10^4 N/m)(0.0036 m^2)
PE = 21.6 J

Added by Jeffrey P.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

11/29/2022

Step 1

Step 1: Recall the formula for elastic potential energy: PE = (1/2)kx^2 Show more…

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A spring with a 1.2x10^4 N/m force constant is stretched 6 cm from its equilibrium position. Calculate the elastic potential energy of this spring. PE = (1/2)kx^2 PE = (1/2)(1.2x10^4 N/m)(0.06 m)^2 PE = (1/2)(1.2x10^4 N/m)(0.0036 m^2) PE = 21.6 J