An object of mass m 1 = 8.30 kg is in equilibrium while connected to a light spr

ID: 1794916 • Letter: A

Question

An object of mass m1 = 8.30 kg is in equilibrium while connected to a light spring of constant k = 100 N/m that is fastened to a wall (see part (a) of the figure below). A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (see part (b) of the figure below). The system is then released and both objects start moving to the right on the frictionless surface.

(a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see the figure above part (c)) and moves to the right with speed v. Determine the value of v.
m/s

(b) How far apart are the objects when the spring is fully stretched for the first time (D in part (d) of the figure above)? (Suggestion: First determine the period of oscillation and the amplitude of the m1-spring system after m2 loses contact with m1.)
cm

Explanation / Answer

(a)

Work = ½kx2 = ½mv2

(Initial KE = 0 J)

½ 100*0.202 = ½ 15.3v2

v = 0.511 m/s

(b)

½ mv2 = TE = ½kA2

½ 8.3(0.511)2 = ½ 100A2

A = 0.1472 m........for m1

Distance traveled by m1

= (k/m)

= (100/8.3)

= 3.47 rad/sec

= 2/T

T = 1.810 sec

x = v (T/4) ..........1/4 of period

x = 0.51(0.4525)

x = 0.2312 m...........for m2

Distance traveled by m2.

so the distance between object when spring is fully streached.

Am1 – xm2 = 0.0837 m = 8.37 cm

negative sign is ignored due to the displacement.

for more explanation..Alternative way.....

initial energy E = Ep = ½kx² = ½ * 100N/m * (0.200m)² = 2 J
At equilibrium, E = Ek = 2J = ½mv² = ½ * (8.30 + 7)kg * v²
v = 0.511 m/s
And now the system "sheds" some Ek in the amount of
Ek = ½ * 7 kg * (0.511 m/s)² = 0.913J,
leaving a system energy of 2J - 0.91J = 1.086J
This results in a spring extension found via
Ep = 1.086 J = ½ * 100N/m * x²
x = 0.1473 m displacement of m1 from equilibrium

How far has m2 traveled? Well,
= sqrt(k/m) = sqrt(100N/m / 8.3kg) = 3.471 rad/s
f = /2 = 0.552/s
so the period T = 1/f = 1.810 s
It takes one fourth of the period for m1 to travel from equilibrium to full extension, or
t = 1.810s / 4 = 0.452s
meanwhile, m2 has traveled
s = v * t = 0.511m/s * 0.452s = 0.231 m

Therefore the separation between m2 and m1 is 0.231m - 0.1473m = 0.0837m = 8.37 cm

Hope it will help you. IF you understand the concept please rate the answer.

(a)