The net force on the block acting perpendicular to the incline is
∑ <em>F</em> = <em>n</em> - <em>w</em> cos(29.4°) = 0
where <em>n</em> is the magnitude of the normal force and <em>w</em> = <em>m g</em> is the weight of the block.
The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.
Solve for <em>n</em> :
<em>n</em> = <em>m g</em> cos(29.4°)
<em>n</em> = (6 kg) (9.80 m/s²) cos(29.4°)
<em>n</em> ≈ 51.2 N
The proton number?
number of protons
The gravitational energy is going up subtracting the energy that was on the ground
