Question

*A block with mass of m is connected to a spring on a ramp (ignore friction) that makes an angle θ with the horizontal. The block is let go while the spring is in its unstretched position. As the block moves on the ramp, its displacement from the unstretched position is x.

Answer 1

`(mg\sin\theta)/(x_0)`

Step-by-step explanation:

At the equilibrium position, we can make the following free body diagram

Where Fn is the normal force, Fs is the force of the spring and mg is the weight of the block.

Then, for the direction parallel to the ramp, we can write the following equation:

`F_(net)=F_s-mg\sin\theta=0`

By Hooke's law, the force of the spring is kx0, where k is the constant of the spring and x0 is the displacement at the equilibrium position. Replacing this expression and solving for k, we get:

`\begin{gathered} kx_0-mg\sin\theta=0 \\ kx_0=mg\sin\theta \\ \\ k=(mg\sin\theta)/(x_0) \end{gathered}`

Therefore, the answer is

`(mg\sin\theta)/(x_0)`