Question

A spring has a force constant k=500 N/m. What is the elastic potential energy stored in it when (a) it is stretched 4.00 cm from equilibrium, (b) it is compressed 3.00 cm from equilibrium, and (c) it is unstretched?

Answer 1

elastic potnetail ennergy U   = 0.5 kx^2

a. when x = 4 cm = 0.04 m

U = 0.5* 500 * 0.04*0.04

U = 0.4 Joules
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when x = 3 cm = 0.03 m

U = 0.5* 500 * 0.03*0.03

U = 0.225 Joules

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c. when un streched x =0

so U = 0 J

Answer 2

we know elastic potential energy = 0.5*K*x*x       where x is stretched

hence

a) PE=0.5*500*0.04*0.04=2/5J

b) PE = 0.5*500*0.03*0.03 = 9/40 J

c)PE=0.5*500*0*0 = 0J

Answer 3

eastic potential energy stored is U= 1/2 k x²

a) U= 0.5*500*16*10*-4

   U= 4000*10^-4 J

b) when streched 3 cm is U= 0.5*500*9*10^-4 J

                                      U=2250*10^-4 J

c) when it is in unstreched position elastic potential energy is zero

Answer 4

Elastic potential energy = 0.5kx^2

k = 500 N/m

(a) x = 0.04m

Energy = 0.5 X 500 X 0.04 X 0.04 = 0.4 J

(b)

x = 0.03m

Energy = 0.5 X 500 X 0.03 X 0.03 = 0.225 J

(c) x = 0

Energy = 0

Answer 5

a. elastic potential energy= (1/2) kx^2=0.5*500*0.04*0.04=0.4J

b. elastic potential energy= (1/2) kx^2=0.5*500*0.04*0.04=0.4J

c. elastic potential energy= 0J

Answer 6

a. elastic potential energy= (1/2) kx^2=0.5*500*0.04*0.04=0.4J

b. elastic potential energy= (1/2) kx^2=0.5*500*0.03*0.03=0.225J

c. elastic potential energy= 0J

Answer 7

Work done to strech the spring a distance x is

Pe(Potential Energy)=1/2kx^2

Convert cm in to m

4cm=0.04 m

Pe=1/2*500*0.04*0.04

a)Pe=0.4 J when spring is streched

b)when spring is compressed

Pe=-1/2 kx^2

Pe=-1/2*500*0.03*0.03

Pe=-0.225J

c) When spring is unstreched

PE_s = 0 when x = 0 (at equilibrium).