Answer:
The wood block reaches a height of 4.249 meters above its starting point.
Explanation:
The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]
[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]
[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)
Where:
[tex]h[/tex] - Maximum height of the wood block, in meters.
[tex]v[/tex] - Initial speed of the block, in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]m[/tex] - Mass, in kilograms.
[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.
[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.
If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:
[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]h = 4.249\,m[/tex]
The wood block reaches a height of 4.249 meters above its starting point.
