Question

Your answer is partially correct. Try again. Two identical 0.12 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 = 0.05 m. Then a constant force of 7 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 xi = 0.02 m and x3 = 0.13 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched. Initial x = 0 m x = x2 Final X = X1 x= x3

Answer 1

x_cm = x₁ + ( x₃ - x₁ / 2 )

= 0.02 + ( 0.13 - 0.02 / 2 )

= 0.02 + 0.055

= 0.075 m

$\Delta$x = 0.075 - x₂ / 2

= 0.075 - 0.05 / 2

$\Delta$x = 0.05 m

a )

d = x₃ - x₂

= 0.13 - 0.05

d = 0.08 m

b )

the work done is $\Delta$ (K_tot + U) = F x d

= 11x 0.08

the change in energy of the extended system is $\Delta$ (K_tot + U) = 0.88 J

c )

(K_tot + U)_final for the extended system is also same = (K_tot + U)_final = 0.88 J

d )

the final translational kinetic energy of the extended system is

K_trans,final = F x $\Delta$ x

= 11 x 0.05

K_trans,final = 0.55 J

e )

v_cm,final = ( 2 x 0.55 / 0.12 + 0.12 )^1/2

v_cm,final = 2.14 m/sec