Answer to Question #98981 in Mechanics | Relativity for Jae

Question

A spring (k=250N/m) keeps a 2.9kg mass stationary on a frictionless 70 degree slope. How much is the string stretched? If mass was moved so that the stretch in the spring was changed to 15cm. What would be the acceleration of the mass?

Answer 1

The Hooke's law state

$$F=k\Delta x$$

The mass-spring equilibrium condition is given by

$$k\Delta x_1=mg\sin\theta$$

So

$$\Delta x_1=\frac{mg\sin\theta}{k}$$

$$=\frac{2.9\:\rm kg\times 9.8\: N/kg\times \sin70^{\circ}}{250\:\rm N/m}=0.107\:\rm m$$

The Newton's second law gives

$$ma=k\Delta x_2$$ $$a=k\Delta x_2/m$$

$$a=250\times (0.15-0.107)/2.9=3.7\:\rm m/s^2$$