Physics Help. Answer please?

[匿名]

2015-12-10 4:57 pm

A 0.110 kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.50 m above the floor. The spring has force constant 2000 N/m and is initially compressed 0.045 m. The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor.

If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

回答 (1)

NCS

2015-12-10 10:44 pm

✔ 最佳答案

Horizontally: spring energy is converted to kinetic energy.
½kx² = ½mv²
½ * 2000N/m * (0.045m)² = ½ * 0.110kg * Vx²
Vx² = 36.8 m²/s²

Vertically: gravitational PE is converted into KE
mgh = ½mv²
Vy² = 2gh = 2 * 9.8m/s² * 1.5m = 29.4 m²/s²

speed = √(Vx² + Vy²) = 8.14 m/s

Hope this helps!

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