A 0.140 kg block of ice is placed against a horizontal, compressed spring mounte
ID: 1540576 • Letter: A
Question
A 0.140 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 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.
A. 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?
Explanation / Answer
when the ice reaches the floor it will have a horizontal component of speed = speed it attains from the stored energy in spring; in addition it will have a vertical component of speed = Vy = 2gh.
Vy = (2)(9.81)(1.10) = 21.582 = 4.646 m/s
SPE = 1/2kx² = (0.5)(2000)(0.045)² = 2.025 J
KE of ice when sliding on table = 1/2mVx² = (0.5)(0.140)Vx² = 0.070Vx²
KE of ice when sliding on table = SPE
0.070Vx² = 2.025
Vx = 2.025/0.070 = 28.928 =5.378 m/s
Speed of block at floor= (Vy²+Vx²)= [(4.646)²+(5.378)²]= 50.5139 = 7.11 m/s ANS
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