Question

a spring of spring constant ( k ) is attached to the bottom of a ramp of negligible friction which makes an angle ( \theta ) with the floor. a block of mass ( m ) is pushed against the spring, compressing it a distance ( x_{0} ), as shown in the figure. the block is then released. what is the distance ( d ) that the block slides up the ramp before momentarily coming to rest?

Explanation:

Step1: Calculate the elastic potential energy

The elastic potential energy of the spring is \(U_{s}=\frac{1}{2}kx_{0}^{2}\).

Step2: Calculate the gravitational potential energy

The gravitational potential energy of the block at the highest point is \(U_{g}=mg(d + x_{0})\sin\theta\).

Step3: Apply the conservation of mechanical energy

Since there is no friction, the elastic potential energy is converted into gravitational potential energy. So \(\frac{1}{2}kx_{0}^{2}=mg(d + x_{0})\sin\theta\).

Step4: Solve for \(d\)

$$
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$$

Answer:

\(d=\frac{kx_{0}^{2}}{2mg\sin\theta}-x_{0}\)