Moving blocks connected by a spring Two identical 0.16 kg blocks (labeled 1 and

Question

Moving blocks connected by a spring Two identical 0.16 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 = 0.04 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 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.

MOTION OF THE CENTER OF MASS OF THE EXTENDED SYSTEM

a. What is the initial location of the center of mass of the extended system? xCM,initial = m

b. What is the final location of the center of mass of the extended system? xCM,final = m

c. How far did the center of mass of the extended system move? xCM = m

POINT PARTICLE SYSTEM

d. What is the initial location of the point particle system? xinitial = m

e. What is the final location of the point particle system? xfinal = m

f. How far did the point particle system move? x = m

g. In the point particle system, what was the distance through which the point of application of the force moved? distance = m

h. How much work was done on the point particle system? W = joules

i. What was the change in the translational kinetic energy of the point particle system? Ktrans = joules

j. What is the final translational kinetic energy of the point particle system? Ktrans = joules

k. What is the mass of the point particle system? m = kg

l. What is the final speed of the point particle system? vcm,final = m/s

EXTENDED SYSTEM

m. The force was applied to a location on the right hand edge of block 2. Look carefully at the diagrams. In the extended system, what was the distance through which the point of application of the force moved? distance = m

n. How much work was done on the extended system? W = joules

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

p. What is (Ktot + U)final for the extended system? (Ktot + U)final = joules

q. What is the final translational kinetic energy of the extended system? Ktrans,final = joules

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

Explanation / Answer

given

0.16 kg

x1 = 0.02 m

x2 = 0.04 m

x3 = 0.12 m

F = constant force of 7 N

a )

X = m1 x1 + m2 x2 / ( m1 + m2 )

X = 0.16 X 0 + 0.16 X 0.04 / ( 0.16 + 0.16 )

X = 0.02 m

b )

X = m1 x1 + m2 x2 / ( m1 + m2 )

X = 0.16 X 0.02 + 0.16 X 0.12 / ( 0.16 + 0.16 )

X = 0.07 m

c )

the center of mass of the extended system move is = 0.02 - 0.07 = 0.05 m

d )

same as the value of center of mass

e )

the center of mass of the extended system move is = 0.02 - 0.07 = 0.05 m

like it is also same

f )

W = F.s

= 7 X 0.05

W = 0.35 J

g )

same as above W = 0.35 J

h )

same as above W = 0.35 J

i )

the value of the mass is = 0.16 + 0.16

= 0.32 kg

j )

V = ( 2 E / m )1/2

V = ( 2 X 0.05 / 0.32 )1/2

V = 0.0559 m/sec

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