Firstly, let's convert all the quantities to the appropriate units in order to carry out the calculations correctly.
The block mass is 190g, which, when converted to kilograms (since 1 kg = 1000 g), gives us 0.190 kg.
The force constant of the spring is given as 1.5 kN/m, which we convert to Newtons per meter (since 1 kN = 1000 N), giving us a spring constant of 1500 N/m.
The compression of the spring is given as 10.0 cm, which we convert to meters (since 1 m = 100 cm), giving 0.10 m.
Furthermore, the incline angle is given as 60 degrees, which should be converted to radians (since 1 degree = π/180 radians; this comes from the fact that the full circle 360 degrees is 2π in radians). This gives us an angle of π/3 radians.
The force that the block exerts on the spring is described by Hooke's Law, which can be expressed as F = kx, where F is the force, k is the spring constant, and x is the spring compression. Substituting the known values, we calculate the force to be 1500 N/m * 0.10 m = 150 N.
The work done on the block by the spring, while it is compressed, is calculated as the area under the force-displacement graph, which is equal to (1/2)Fx. Therefore, by substituting the known values, we calculate the work done to be (1/2) * 150 N * 0.10 m = 7.5 J.
This work done is then converted into potential energy as the block is released and moves up the inclined plane. The increase in potential energy for the block is given by ΔPE = mgh, where m is the mass of the block, g is the acceleration due to gravity (which is 9.8 m/s² on the surface of the Earth), and h is the increase in height.
Since the work done on the block is equal to the block's increase in potential energy (since no other forces do work on the block), we can equate these two quantities and solve for h, the increase in height:
7.5 J = 0.190 kg * 9.8 m/s² * h
Solving for h, we find that the block rises to a height of approximately 4.03 m up the incline.
Now we have two results:
- The work done on the block by the spring is 7.5 J.
- The block rises to a height of approximately 4.03 m up the incline.
These are the final answers for this problem.