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

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

Step-by-step 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.

`(1)/(2)\cdot m \cdot v^(2) = m \cdot g\cdot h + \mu_(k)\cdot m\cdot g\cdot s \cdot \sin \theta`

`(1)/(2)\cdot v^(2) = g\cdot h + \mu_(k)\cdot g\cdot \left((h)/(\sin \theta) \right)\cdot \sin \theta`

`(1)/(2)\cdot v^(2) = g\cdot h +\mu_(k)\cdot g\cdot h`

`(1)/(2)\cdot v^(2) = (1 +\mu_(k))\cdot g\cdot h`

`h = (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\,(m)/(s)`,
`\mu_(k) = 0.20` and
`g = 9.807\,(m)/(s^(2))`, then the height reached by the block above its starting point is:

`h = (\left(10\,(m)/(s) \right)^(2))/(2\cdot (1+0.20)\cdot \left(9.807\,(m)/(s^(2)) \right))`

`h = 4.249\,m`

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