A box is attached to a spring with a spring constant of 350 N/m on a frictionless incline that makes an angle of 30° with respect to horizontal as shown. The box is at rest when the spring is stretched 0.15 meters from its equilibrium length.

What is the mass of the box?

Question

A box is attached to a spring with a spring constant of 350 N/m on a frictionless incline that makes an angle of 30° with respect to horizontal as shown. The box is at rest when the spring is stretched 0.15 meters from its equilibrium length.

What is the mass of the box?

anonymous · Answer

attachment

jamesj · Answer

The box is at rest when the net forces on the box are zero.  Clearly there is a spring force,
$$ F_s = -kx = -(350)(0.15) = ... $$
and that must be exactly countering the component of the gravitational force in the direction of the incline, the direction in which the spring force $ F_s $ is acting on the box.  

Find an expression for that component of the gravitational force, set it equal to $ F_s $ and solve for the mass $ m $ of the box.

anonymous · Answer

Ok so I solved for Fs and got -52.5. Then using:
 $$F=ma$$

I did -52.5/-9.81 and got the mass to be 5.35kg. It says the correct answer should be 10.7kg though

jamesj · Answer

No.  What you haven't done is found the component of the gravitational force that is in the direction of the plane.  You have taken the ENTIRE gravitational force and matched it with the spring force.  But this is clearly wrong.  You only want that component that is in the direction of the incline; i.e., parallel to the spring force.

anonymous · Answer

ok i think I got it. thanks!