Question

A 4.0 kilogram object is dropped from a height of 1.0 meter onto a spring with a spring constant of 850 newtons per meter. Approximately what distance will the spring compress when the object lands on it? (1) 3.0 dm
(2) 9.0 cm

(3) 9.0 mm

(4) 2.0 dm

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Answer 1

Answer:

(1) 3.0 dm

Explanation:

Due to the law of conservation of energy, the initial gravitational potential energy of the object will be all converted into elastic potential energy of the spring as the object hits the spring:

$mgh = \frac{1}{2}kx^2$

where

m = 4.0 kg is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.0 m is the initial height of the object

k = 850 N/m is the spring constant

x is the compression of the spring

Solving the equation for x, we find:

$x=\sqrt{\frac{2mgh}{k}}=\sqrt{\frac{2(4.0 kg)(9.8 m/s^2)(1.0 m)}{850 N/m}}=0.30 m=3.0 dm$