Question

2. Calculate the force of elasticity (or tension) with which an 8 kg hanging candlestick pulls the cord. How much is the deformation of the cord when it is known that the coefficient of elasticity has values ​​k = 4 × 10⁴ N / m.

Answer 1

(a) We can find the force of tension from the Newton's Second Law of Motion:

$$\sum F_y=ma_y=0,$$ $$T-mg=0,$$ $$T=mg=8\ kg\cdot 9.8\ \dfrac{m}{s^2}=78.4\ N.$$

(b) Finally, we can find the deformation of the cord from the Hooke's Law:

$$T=F=kx,$$ $$x=\dfrac{T}{k}=\dfrac{78.4\ N}{4\cdot10^4\ \dfrac{N}{m}}=19.6\cdot10^{-4}\ m.$$

Answer:

(a) $T=78.4\ N.$

(b) $x=19.6\cdot10^{-4}\ m.$