Answer to Question #94047 in Mechanics | Relativity for Mads.

Question

The displacement of a block attached to a horizontal spring whose spring constant is 2N/m is given by

x=1.8cos(6.6t−0.8)m.
The acceleration of the block at t=1s is
m/s2.

Answer 1

Solution:

The velocity of a block:

$\upsilon =\frac{dx}{dt}=1.8\cdot{6.6}\cdot{(-sin(6.6t-0.8))}=$

$=-11.9(sin(6.6t-0.8))$

The  acceleration of a block:

$a =\frac{d\upsilon}{dt}=-11.9\cdot{6.6}\cdot{(cos(6.6t-0.8))}=$

$=-78.5(cos(6.6t-0.8))$

The acceleration of the block at t=1s:

$a(1)=-78.5(cos(6.6-0.8))=-78.5(cos(5.8))=$

$-78.5\cdot{0.9949}=-78.1$ m/s²

Answer:

The acceleration of the block at t=1s is -78.1 m/s².