Kinetic energy on an incline

A mass m=5 kg starts with zero velocity at time t=0s on a plane inclined at an angle β with respect to the horizontal. We set g=10 ms-2 and sin(β)=1/2 and we consider that there is no friction. After traveling a distance of 20 m, its kinetic energy K is:

Answer 500J

I found 490J, not sure what is wrong with my work, i did estimate the value of square root of 2

F = mgsinβ so a = gsinβ = 5m/s²
v² = u²+2as = 10 * square root of 2 = 14
Ek = 1/2mv² = 490J

hello! here I used a different energy conservation method and calculated the velocity from potential energy= kinetic energy at the peak height. I hope it makes sense.