Physics, asked by krishnagupta6464, 1 year ago

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

Answered by Anonymous
10

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

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