Can someone please help me I'm stuck on this question?

Question

A 1.20-kg object is held 1.00 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is dropped onto the spring.

(a) How far does the object compress the spring?

_____ m

(b) Repeat part (a), but this time assume a constant air-resistance force of 0.600 N acts on the object during its motion.

_____ m

(c) How far does the object compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s² and air resistance is neglected?

_____ m

Answer 1

Let

m = 1.2 kg is the mass of the object,

h = 1 m is the initial height of the object,

k = 315 N/m is the spring's force constant,

g is gravitational acceleration: 9.8 m/s^2 for (a), (b) and 1.63 m/s^2 for (c)

F_r is the force of resistance: 0 for (a), (c) and 0.6N for (b)

x is the maximum compression of the spring to be calculated

We calculate x from conservation of energy.

All the potential energy E_p the object has at the beginning gets

converted into the spring energy E_s and the work W_r of the resistance.

The potential energy E_p is with respect to the point x of maximum compression,

because at that point all the energy conversion has finished.

E_p = (h + x)mg

The spring energy is given by

E_s = kx²/2

The work of the resistance is the product of the resistance force over the distance travelled.

W_r = F_rx

Conservation of energy:

E_p = E_s + W_r

Substituting

(h + x)mg = kx²/2 + F_rx

This is a quadratic equation with known coefficients,

which I hope you can solve for the three cases (a), (b), (c).

If you have trouble with the equation, let me know.

Answer 2

a)

Let A be initial point, Let B be final point when spring is fully compressed

Initial Velocity at A = 0

Final Velocity at maximum compression at B = 0

Therefore net kinetic energy is zero

-(1/2) * k * (Xb^2-Xa^2) - m * g * (hb-ha) = (1/2) m * (Vb^2-Va^2)

-(1/2) * k * (Xb^2-Xa^2) - m * g * (hb-ha) = 0

Xb = ?

Xa = 0

hb = 0

ha = 1 m

-(1/2) * k * (Xb^2) - m * g * (-ha) = 0

mgha = (1/2) * k * (Xb^2)

2/k * mgha = Xb^2

2/315 * 1.20 * 9.81 * 1 = Xb^2

Xb^2 = 0.0747

Xb = 0.273 m

(b)

Resistive force creates negative work. Since it opposes the motion

-(1/2) * k * (Xb^2) - m * g * (-ha) -Fr * ha = 0

mgha-Frha = (1/2) * k * (Xb^2)

ha * (mg-Fr) = (1/2)*k*(Xb^2)

(2/k) *ha * (mg-Fr)= Xb^2

Xb = 0.266 m

Sanity check, less compression due to resistive force

(c) Take formulation from part a and change the g factor

2/k * mgha = Xb^2

2/315 * 1.20 * 1.63 * 1 = Xb^2

Xb = 0.111 m