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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. The 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

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LammettHash

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 PE :

PE = m g h

PE = (2.0 kg) (9.8 m/s²) (3.0 m)

PE = 58.8 J

Energy is conserved throughout the block's descent, so that PE at the top of the curve is equal to kinetic energy KE at the bottom. Solve for the velocity v :

PE = KE

58.8 J = 1/2 m v ²

117.6 J = (2.0 kg) v ²

v = √((117.6 J) / (2.0 kg))

v ≈ 7.668 m/s ≈ 7.7 m/s

The velocity of the block at maximum kinetic energy is 7.668 m/s.

From  potential energy formula,

PE = m g h

Where,

m = mass = 2 kg

g -gravitational acceleration = 9.8 m/s²

h - height = 3 m

Put the values in the formula,

PE = (2.0 kg) (9.8 m/s²) (3.0 m)  

PE = 58.8 J

Since,  

PE = KE

Thus,

58.8 J = 1/2 m v ²  

117.6 J = (2.0 kg) v ²  

v = √((117.6 J) / (2.0 kg))  

v = 7.668 m/s

Therefore, the velocity of the block at maximum kinetic energy is 7.668 m/s.

To know more about  potential energy

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