A 20-N/cm spring is stretched on a frictionless incline of 30 degrees by fixing its upper end and connecting a 4.0-kg box of bananas to its lower end. The length between its coils is 18 cm in this stretched position, when the system is in static equilibrium. What is the natural length of the spring?

Question

A 20-N/cm spring is stretched on a frictionless incline of 30 degrees by fixing its upper end and connecting a 4.0-kg box of bananas to its lower end.  The length between its coils is 18 cm in this stretched position, when the system is in static equilibrium.  What is the natural length of the spring?

anonymous · Answer

By F=ma, 
along the incline, because it is in static equilibrium
 $$\sum F= F_s - F_g =0$$
$$F_s =F_g$$
$$ke=mg \sin 30^o$$, where e=18-x(extension of the spring, original length x)
$$20(18-x)=4 (10)(\sin 30^o)$$
Solving for x, we get x=17cm.

anonymous · Answer

note that my main system is the box of bananas.