A 2.0 kg wood block is launched up a wooden ramp

[bobbarkernar]

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30 degree angle. The block's initial speed is 10 m/s. The coefficient of kinetic friction of wood on wood is .200

1)What vertical height does the block reach above its starting point?
2)What speed does it have when it slides back down to its starting point?

can someone please help me

 

[bobbarkernar]

i found the x and y component of the weight:
wx=196sin(30) and wy is 196cos(30)

im not sure where to go from here

 

[kyle8921]

1) To determine the vertical height reached by the block, we can use the conservation of energy principle. At the starting point, the block has only kinetic energy due to its initial speed, and at the highest point, it has only potential energy due to its height. Therefore, we can equate the initial kinetic energy to the final potential energy.

Initial kinetic energy = Final potential energy
1/2 * mass * velocity^2 = mass * gravity * height
1/2 * 2kg * (10m/s)^2 = 2kg * 9.8m/s^2 * height
100J = 19.6m/s * height
height = 5.1m

Therefore, the block reaches a vertical height of 5.1 meters above its starting point.

2) To determine the speed of the block when it slides back down to its starting point, we can use the equation for final velocity in a one-dimensional motion with constant acceleration:

Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance
Final velocity^2 = 10m/s^2 + 2 * (-9.8m/s^2) * 5.1m
Final velocity^2 = 10m/s^2 - 100m/s^2
Final velocity^2 = -90m/s^2
Final velocity = √(-90m/s^2)
Final velocity = 9.5m/s

Therefore, the block will have a speed of 9.5 meters per second when it slides back down to its starting point.