Question

A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0650 m. HINT (a) What is the elastic potential energy of the block-spring system (in J)? (b) If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring. n/s

Answer 1

Answer:

a) The initial elastic potential energy of the block-spring system is 28.113 joules.

b) The final speed of the block is approximately 4.192 meters per second.

Explanation:

a) By applying Hooke's law and definition of work, we define the elastic potential energy ($U_{g}$), measured in joules, by the following formula:

$U_{g} = \frac{1}{2}\cdot k\cdot x^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$x$ - Deformation of the spring, measured in meters.

If we know that $k = 865\,\frac{N}{m}$ and $x = 0.065\,m$, then the elastic potential energy is:

$U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)$

$U_{g} = 28.113\,J$

The initial elastic potential energy of the block-spring system is 28.113 joules.

b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:

$U_{g} = \frac{1}{2}\cdot m\cdot v^{2}$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Final speed, measured in meters per second.

Then, the final speed is cleared:

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

If we know that $U_{g} = 28.113\,J$ and $m = 3.20\,kg$, then the final speed of the block is:

$v = \sqrt{\frac{2\cdot (28.113\,J)}{3.20\,kg} }$

$v \approx 4.192\,\frac{m}{s}$

The final speed of the block is approximately 4.192 meters per second.