Answer:
The slope of a position-time graph can be calculated as:

where
is the increment in the y-variable
is the increment in the x-variable
We can verify that the slope of this graph is actually equal to the velocity. In fact:
corresponds to the change in position, so it is the displacement, 
corresponds to the change in time
, so the time interval
Therefore the slope of the graph is equal to

which corresponds to the definition of velocity.
Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
The formula that is usually used for the calculation of power is the product of force applied and the speed at which the action is done. That is,
P = Fv
We let d be the distance covered and the equation for power would be,
P = (500 N)(d/240 s)
P = 2.08d