The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :
<em>PE</em> = <em>m g h</em>
<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)
<em>PE</em> = 58.8 J
Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :
<em>PE</em> = <em>KE</em>
58.8 J = 1/2 <em>m v</em> ²
117.6 J = (2.0 kg) <em>v</em> ²
<em>v</em> = √((117.6 J) / (2.0 kg))
<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s