Moving blocks connected by a spring Two identical 014 kg blo

Moving blocks connected by a spring
Two identical 0.14 kg blocks (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper diagram, where x₂ = 0.06 m. Then a constant force of 8 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower diagram, where x₁ = 0.02 m and x₃ =0.12 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched.

What is the change in energy of the extended system?
(K_tot + U) =  joules

What is (K_tot + U)_final for the extended system?
(K_tot + U)_final =  joules

What is the final translational kinetic energy of the extended system?
K_trans,final =  joules

What is the final speed of the center of mass of the extended system?
v_cm,final =  m/s

What is the final vibrational energy of the extended system (spring potential energy plus kinetic energy relative to the center of mass)?
(U + K_rel)_final =  joules

Solution

a)

Initial center of mass

X_CM,initial=M₁X+M₂X₂/(M₁+M₂)

X_Cm,initial =0.14*0+0.14*0.06/(0.14+0.04)

X_cm,initial=0.03 m

b)

Final Center of mass

X_Cm,final=M₁X₁+M₂X₃/(M₁+M₂)

X_cm,final=(0.14*0.02+0.14*0.12)/(0.14+0.14)

X_cm,final=0.07 m

c)

dX_Cm =0.07-0.03 =0.04 m

d)

X_initial=0.03 m

e)

X_final=0.07 m

f)

dX=0.04 m

g)

Distance =0.04 m

h)

Workdone

W=Fd =8*0.04

W=0.32 J

i)

Translational Kinetic energy

j)

dKE_trans=0.32 J

k)

K_trans=0.32 J

l)

mass of point particle system

M=2*0.14=0.28 kg