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
Yes, your calculations for both questions are correct. Good job!
To explain how you found the answers:
1) To find the change in potential energy of the system, you used the formula for potential energy stored in a spring: PE = (1/2)kx^2, where k is the spring constant and x is the displacement or stretch from the equilibrium position. Plugging in the values, you correctly calculated the change in potential energy to be 1.8 Joules.
2) To calculate the work done by the spring, you used the formula for work done by a spring: W = (1/2)kx^2(final) - (1/2)kx^2(initial), where k is the spring constant and x(final) and x(initial) are the final and initial displacements or stretches, respectively. Plugging in the values, you correctly determined that the work done by the spring is 1.6 Joules.
Keep up the good work! If you have any more questions, feel free to ask.