Answer:
m = 1.35 kg
Explanation:
- In absence of friction, total mechanical energy must be conserved, so the sum of the initial elastic potential energy plus the initial kinetic energy, must be equal to the sum of the final kinetic energy plus the final elastic potential energy, as follows:
[tex]U_{i} + K_{i} = U_{f} + K_{f} (1)[/tex]
- As the block starts from rest, this means that Ki =0.
- We can express the elastic potential energy at any point , as follows:
[tex]U = \frac{1}{2} * k * \Delta x^{2} (2)[/tex]
where k is the spring constant = 60.0 N/m, and Δx is the distance from
the equilibrium position.
- Replacing Ui, Uf and Kf in (1), we have:
[tex]\frac{1}{2} * k *\Delta x_{i} ^{2} =\frac{1}{2} * k *\Delta x_{f} ^{2} +\frac{1}{2} * m *v ^{2} (3)[/tex]
- Replacing by the givens, and solving for m, we get:
m =1.35 kg
