Question

a mass of 4 kg rests on a smooth plane inclined at 30 degrees to the horizontal. it is held in equilibrium by a light elastatic string attached to the mass on to a point on the plane. find the extension in the string if it is known that a force of would double the natural lenght of 1.25 meters

Answer 1

We have

$$m=4 (kg); \Delta l_0=1,25 (m); F=49 (N); \alpha=30; g=10 (m/c^2)$$

Draw a scheme

stiffness coefficient

$$k=\frac{F}{\Delta l_0}$$

Newton's law

$$x: mgsin(\alpha)-k\Delta l=0$$

And we have

$$\Delta l=\frac{mgsin(\alpha)\Delta l_0}{F}$$

$$\Delta l=\frac{4\cdot 10\cdot sin(30) \cdot 1,25}{49}=0.51 (m)$$