A 35.0 kg mass is suspended at equilibrium on a spring that has a spring constant of 1.23*10^3 N/m. The mass is pulled down by 18.5 cm... from its equilibriumÂ position, and then released. When the mass has risen up by a distance equal to 27.0 cm after being released, what is its acceleration?
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Tushar Chandra
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The spring has a spring constant of 1.23*10^3 N/m. A 35 kg mass is suspended from it. The mass is pulled down so that the spring is extended by 18.5 cm. It is then released and the rises up by 27 cm.
After the mass has risen 27 cm it is at a height equal to 27 - 18.5 = 8.5 cm above the equilibrium position of the spring. There are two forces acting on the mass here:
- A force due to the gravitational force of attraction of the Earth which accelerates it downwards by 9.8 m/s^2
- A force exerted by the spring downwards which is equal to 1.23*10^3*8.5*10^-2 = 104.55 N. This accelerates the spring downwards by 104.55/35 = 2.987 m/s^2
The net acceleration of the body is 9.8 + 2.987 = 12.787 m/s^2.
At the required position the acceleration of the body is 12.787 m/s^2 downwards.
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