Dual Springs

Classical Mechanics Level 3

A block of mass M = 3 kg M=3\text{ kg} is initially in equilibrium and hanging by 2 identical springs as shown in figure above. If one of the springs is cut from the lower point at t = 0 t=0 , then find acceleration of block in m/s 2 \text{m/s}^2 at t = 0 t=0 .

Details and Assumptions :

  • Do not consider rotational effects of the block.

  • Take g = − 10 m/s 2 g=-10\text{ m/s}^2 .

  • Take upward as positive direction

− 10 m/s 2 -10 \text{ m/s}^2 + 5 m/s 2 +5 \text{ m/s}^2 − 5 m/s 2 -5 \text{ m/s}^2 − 20 m/s 2 -20 \text{ m/s}^2

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Jus Jaisinghani by Jus Jaisinghani , 21, India

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1 solution

Kai Ott
Jul 18, 2016

Relevant wiki: Newton's Laws of Motion

Gravitational force = F G = m × g = 30 N F_{G} = m × g = 30N .

Spring force (of one spring) = F S F_{S} .

Acceleration = m a = 2 F S − F G ma = 2F_S - F_G .

State of equilibrium (before the spring is cut): 0 = 2 F S − F G = 0 → F S = F G 2 0= 2 F_S - F_G = 0 \rightarrow F_S = \frac{F_G}{2} .

When one spring is cut: m a = F S − F G = − F G 2 = − m g 2 ma=F_S - F_G = -\frac{F_G}{2} = -m\frac{g}{2} .

a = − 5 m s 2 a = -5 \frac{m}{s^2} .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice, one my mistake was I took only 1 spring force instead of two

Arnav Das - 4 years, 11 months ago

good solution

Jus Jaisinghani - 4 years, 11 months ago

Me too...simple but great!!!

Javier Martínez Dominguez - 4 years, 11 months ago

I forgot consider the equilibrium state first...

Javier Martínez Dominguez - 4 years, 11 months ago

I think in 2nd case equation should be Fg-Fs=ma. As the object Wil move downward after cutting the spring not upward......please correct my doubt.

Anshul Sanghi - 4 years, 10 months ago

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The acceleration is downwards,you are correct.The sign conventions are now mentioned in the problem please recheck n ask if not understood

Jus Jaisinghani - 4 years, 10 months ago

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