A 0.600 kg block (m) is dropped from a height of 16.0 m (h) onto a vertically mo

Question

A 0.600 kg block (m) is dropped from a height of 16.0 m (h) onto a vertically mounted spring device with a horizontal board upon which the block will land as shown in the figure. The spring, with a spring constant of 6000 N/m, is initially uncompressed. Use conservation of energy to solve the following:

(a) What will be the velocity (v) of the block as it makes contact with the board? Assume that air resistance can be ignored, and be sure to use energy and not kinematics to solve this problem.

(b) To what maximum distance (d) will the spring be compressed? Be sure to consider all changes to potential energy.

Explanation / Answer

1. mgh = 1/2mv^2

=> v = sqrt(2gh)

=> v = sqrt(2*9.8*16) = 17.70875 m/s

2. F*x = 1/2*m*v^3

=> x = (1/2)(m*v^2/F

=> x = (1/2)*0.6*313.6/6000 = 0.01568 mt. = 1.568 cm

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