An object of mass m_1 = 9.90 kg is in equilibrium when connected to a light spri

Question

An object of mass m_1 = 9.90 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in part [a] of the figure below. A second object, m_2 = 7.00 kg, is slowly pushed up against m_1, compressing the spring by the amount A = 0.170 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 m_1 reaches the equilibrium point, m_2 loses contact with m_1 (see part [c] in the above figure) 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 (the distance D in part (d) of the above figure)? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm

Explanation / Answer

When the spring and masses return to the equilibrium point
The masses will have KE of
.5*(m1+m2)*v^2
and the spring will have lost PE of
.5*k*x^2
or
.5*100*0.17^2

set up the equation
.5*(9.9+7.0)*v^2=.5*100*0.17^2
solve for v
v=.412

b,

the period will determine when the spring reaches max extension. Then compute the distance that the m2 has traveled over a period of 1/4 of the total oscillation
for this part
=sqrt(k/m)
in rad/s
the first time the spring reaches full extension after m2 is launched is
*t=2*pi/4
or
sqrt(k/m)*t=pi/2
solve for t

t=sqrt(m/k)*pi/2
then the amplitude is truncated from the initial release A because of the loss of m2
using energy
.5*m1*v^2=.5*k*x^2
solve for x
x=v*sqrt(m1/k)
that is the distance m1 will travel to the full extension of the spring
plug in the numbers
x=0.412*sqrt(9.9/100)
x=0.129m
now compute the distance that m2 has traveled

x2=0129*3.1...
0.402 m
the distance they are apart is
x2-x
or
d=0.402-0.129
0.22 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.