Question

A 0.200 kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.10 m above the floor. The spring has force constant 1950 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?

Answer 1

given

m = 0.200 kg block of ice

h = 1.10 m above the floor

k = force constant 1950 N/m

x = compressed 0.045 m.

v = ?

using conservation of energy we get,

1/2 m v² = 1/2 k x² + m g h

0.5 x 0.2 x v² = 0.5 x 1950 x 0.045² + 0.2 x 9.8 x 1.1

0.1 x v² = 1.9743 + 2.156

0.1 x v² = 4.1303

v² = 4.1303 / 0.1

= 41.303

so the speed of the block of ice when it reaches the floor is v = 6.4267 m/sec