Question

A block of mass m= 0.9 kg is attached to a spring of spring constant k= 500 N/m. the block is pulled to an initial position x = 5 cm to the right of the equilibrium point as shown below, and released from rest. Find the speed of the block as it passes through equilibrium if the coefficient of kinetic friction between the block and the surface μ_k = 0.35.

Answer 1

$mx''+kx=-F_{fr}$

$F_{fr}=\mu_kmg=0.35\cdot0.9g=3.1$ N

then:

$0.9x''+500x=-3.1$

characteristic equation:

$0.9r^2+500=0$

$r=\pm 23.6i$

complementary solution:

$x_c=c_1cos23.6t+c_2sin23.6t$

particular solution:

$x_p=A$

$500A=-3.1 \implies A=-0.0062$

so,

$x(t)=c_1cos23.6t+c_2sin23.6t-0.0062$

we have:

$x(0)=0.05$

$x'(0)=0$

then:

$c_1=0.05-0.0062=0.04$

$x'(t)=-23.6c_1sin23.6t+23.6c_2cos23.6t$

$x'(0)=23.6c_2=0\implies c_2=0$

$x(t)=0.04cos23.6t-0.0062$

$x'(t)=-0.94sin23.6t$

for the speed of the block as it passes through equilibrium:

$x(t)=0.04cos23.6t-0.0062=0$

$t=arccos(0.0062/0.04)/23.6=0.06$ s

$x'(0.06)=v(0.06)=-0.94sin(23.6\cdot0.06)=-0.93$ m/s