Solve question 5on elasticity

Natural length, $L_0 = 0.8$ m
Modulus of elasticity, $\lambda = 100$ N
Extended length, $L = 1.2$ m

The extension is the difference between the extended length and the natural length.

Hooke's Law states that $T = \frac{\lambda e}{L_0}$.

The tension in the string is $\boxed{\text{50 N}}$.

Natural length, $L_0 = 0.8$ m
Modulus of elasticity, $\lambda = 100$ N
Tension, $T = 150$ N

Rearrange Hooke's Law ($T = \frac{\lambda e}{L_0}$) to solve for $e$:

The extended length is the natural length plus the extension.

The length of the string when the tension is 150 N is $\boxed{\text{2.0 m}}$.