Answer to Question #124081 in Mechanical Engineering for Emmanuel Kojo Mensah Kent

Question

An object of mass 0.5 kg is attached to an elastic string with natural length 1.2 m and
causes an extension of 8 cm when the system hangs vertically in equilibrium.
(i) What is the tension in the spring?
(ii) What is the modulus of elasticity of the spring?
(iii) What is the mass of an object which causes an extension of 10 cm?

Answer 1

(i) According second Newton's Law in equilibrium

$$\vec{T}+m\vec{g}=0$$

$$\vec{T}=-m\vec{g}$$

$$|\vec{T}|=0.5kg\cdot9.8m/s^2=4.9N$$

(ii) According Hooke's Law

$$T=-kx$$

$$k={4.9N\over 0.08m}=61.25N/m$$

(iii)

$$m_1g=kx_1$$

$$m_1={kx_1\over g}$$

$$m_1={61.25N/m\cdot0.1m\over 9.8m/s^2}=0.625kg$$

An object of mass 0.625 kg causes an extension of 10 cm when the system hangs vertically in equilibrium.