Question

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vertical height does the block reach above its starting point? Use the coefficients μk=0.20 andμs=0.50.

Answer 1

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.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$, $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$, then the height reached by the block above its starting point is:

$h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 4.249\,m$

The wood block reaches a height of 4.249 meters above its starting point.