A mass is attached to a spring is 0.2 kg. The spring's constant is 40 N/m. The mass is stretched to a position of 0.3 m and released. The final position is 0.1 m. #1) Find the change in potential energy of the system. #2) What is the work done by the spring?
#1) Change in P.E.=(1/2)(40 N/m)(0.3 m)^2
" "= 1.8 Joules
#2)W=[1/2(40N/m).3m^2]-[1/2(40N/m).1m^2]
W= 1.6 Joules
Can you please tell me if I am correct. I thank you in advance for your time. JL
2 answers