Question

A mass is attached to a spring. The rest position of the spring is 45 cm. The mass is hen pulled to stretch the spring to a position of 51 cm. If the spring constant is 805 N/m.
What is the elastic potential energy in this spring at this point?
Answer:
5) What is the first law of Thermodynamics?
Answer:
6) What formula would you use for law of conservation of mechanical energy?
Answer:
7) What equations could be included in a problem with law of conservation of energy if there is a spring shooting an object that is above zero meters

Answer 1

1) Elastic potential energy:
x = 51 cm − 45 cm = 6.0 cm = 0.06 m.
U = 1/2 k x^2 = 0.5(805 N/m)(0.06 m)^2 = 1.45 J (≈ 1.4 J).

5) First law of thermodynamics:
Energy is conserved. For a closed system: ΔU = Q − W,
where ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system. (Equivalent statements: energy cannot be created or destroyed; total energy change equals heat added minus work done.)

6) Formula for conservation of mechanical energy:
KE_i + PE_i = KE_f + PE_f,
i.e. 1/2 m v_i^2 + PE_i = 1/2 m v_f^2 + PE_f, when non-conservative forces (friction, etc.) do no net work.

7) Equations to include for a spring shooting an object above zero meters:
- Spring potential: U_s = 1/2 k x^2 (x measured from spring equilibrium).
- Gravitational potential: U_g = m g h (choose h = 0 reference).
- Kinetic energy: KE = 1/2 m v^2.
- General energy balance:
1/2 k x_i^2 + m g h_i + 1/2 m v_i^2 = 1/2 k x_f^2 + m g h_f + 1/2 m v_f^2.
Common special cases:
- Spring energy all converted to grav. + kinetic (object launched from rest, spring returns to equilibrium): 1/2 k x^2 = m g h + 1/2 m v^2.
- Maximum height (v = 0): h_max = (1/2 k x^2 − 1/2 k x_f^2) / (m g). (If spring ends uncompressed, second term = 0.)
Include any non-conservative work W_nc on the right if there are losses: initial energies − W_nc = final energies.