A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.
If the block is released from rest at a distance of 2.80 m above the floor, what is its speed just before it strikes the floor
1 answer:
Answer: 7.41 m/s
Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.
Kinetic energy = mv²/2, potential energy = mgh
mv²/2 = mgh
v²/2 = gh
v² = 2gh
v = √2gh
Where g = 9.8 m/s², h = 2.80m
v = √2×9.8×2.8 = 7.41 m/s
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