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.
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