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son4ous [18]
3 years ago
13

A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T

he block then slides onto a horizontal plane where it finally comes to rest a distance D from the beginning of the plane. The curved incline is frictionless, but there is a coefficient of friction, μ = 0.5, on the horizontal plane. Determine the velocity of the box at maximum kinetic energy
Physics
1 answer:
Zigmanuir [339]3 years ago
3 0

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

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