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A 0.1-kilogram block is attached to an initially unstretched spring of force constant k = 40 newtons per meter as shown above. The block is released from rest at time t=0.

What is the amplitude, in meters, of the resulting simple harmonic motion of the block?

A) 1/40m
B) 1/20m
C) 1/4m
D) 1/2m

Respuesta :

sitiadriani

Answer:

The amplitude, in meters, of the resulting simple harmonic motion of the block 1/40 m

Explanation:

given information:

the block mass, m = 0.1 kg

spring of force constant, k = 40 N/m

release at t = 0

the formula for simple harmonic motion is

F = kx

where

F = force (N)

k = spring of force constant (N/m)

x = amplitude (m)

thus,

F = kx

x = F/k

  = m g/k

  = 0.1 x 10/40

  = 1/40 m

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