Question

A spring is hung from the ceiling. When a 0.450 kg block is attached to the free end of the spring and released from rest, the block drops 5.0 cm before momentarily coming to rest, after which it moves back upward. What is the spring constant of the spring?

Answer 1

The spring constant of the spring is determined using Hooke's Law, by equating the spring force to the block's weight at the point of maximum displacement. The calculation results in a spring constant of 88.3 N/m for the given mass and displacement.

Step-by-step explanation:

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:

F = -kx

In this scenario, when the block comes to a momentary rest, the spring force is equal to the weight of the block. The block's weight (W) can be calculated by:

W = mg

Where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s2). The displacement (x) is 5.0 cm or 0.050 m.

The weight of the block is:

W = 0.450 kg × 9.81 m/s2 = 4.415 N

Setting the magnitude of the weight equal to the spring force gives us:

kx = mg

Solving for k, we find:

k = mg/x

k = 4.415 N / 0.050 m = 88.3 N/m

Therefore, the spring constant of the spring is 88.3 N/m.

Answer 2

88.2 N/m

set up equation

when the block drops to 5.0 cm, the weight of block and the spring force on the block are equal because there is no acceleration on the block (at rest)

so mg = Fs

due to Hooke's law, Fs = kx

where k is the spring constant and x is the displacement

mg = Fs = kx

mg = kx

values

convert given values to standard units

m = 0.450 kg

g = 9.8 m/s^2

x = 5.0 cm = 0.05 m

plug in values and solve

mg = kx

0.45 * 9.8 = 0.05x

x = 0.45 * 9.8 / 0.05

x = 88.2 N/m