Solve this:
Q8. The spring mass system shown in the figure is in equilibrium at rest. If the mass is pushed up by a distance mg/2k and released, its instantaneous acceleration will be
(A) g
(B) 2g
(C) g/2
(D) zero
Q9. Find the acceleration of block of mass m. Assume pulleys are massless and frictionless.
(A) g/3
(B) 2g/3
(C) g/2
(D) None of these
Answers
8) At equilibrium, mg=kx => x=mg/k; (this is the extension in the spring at equilibrium)
It’s given that the extension is reduced by a distance = mg/2k => new extension in the spring = x-mg/2k = x/2;
Equation of motion at the instant the mass is released from x/2 position
mg-kx/2=ma =>mg-mg/2=ma=> a = g/2
All the best.
9)T - m1g = m1a ...........1
m2g - T = m2a ...........2 (m2 > m1 )
adding both
a = (m2-m1)g/(m1+m2) .............3
now , m1 will move upward but m2 will move downward ...
accleration of m1 is a1 = a(j) (upward is taken as +ve)
accleration of m2 is a2 = a(-j) (downward is taken as -ve)
accleration of center of mass is acom = m1a1 + m2a2 /m1+m2
acom = (m1a - m2a)/m1+m2 (j)
substituting a from eq 3
acom = -g(m2-m1)2/(m1+m2)2 (j)
accleration of com will be in downward direction