Answer
Ans:
$$
s_W=2.89 \mathrm{~m}
$$
Work Step by Step
Angular displacedment of gear $A$ at $t=5 \mathrm{~s}$ is:
$$
\begin{gathered}
d \theta=\omega d t \\
\int_0^{\theta_A} d \theta=\int_0^{5 s} 100(4+t) d t \\
\theta_A=3250 \mathrm{rad}
\end{gathered}
$$
And $r_A \theta_A=r_B \theta_B$, then:
$$
\theta_B=\frac{r_A}{r_B} \theta_A=\left(\frac{40}{225}\right)(3250)=577.78 \mathrm{rad}
$$
Since gear $C$ is also attached to the same shaft as gear $B$
$$
\theta_C=\theta_B
$$
Also $\theta_D r_D=\theta_C r_C$
$$
\theta_D=\frac{r_C}{r_D} \theta_C=\left(\frac{30}{300}\right)(577.78)=57.78 \mathrm{rad}
$$
Since shaft $E$ is attached to gear $D$
$$
\theta_E=\theta_D
$$
Then the distance at which load $W$ is lifted is?
$$
s_W=r_E \theta_E=(0.05)(57.78)=2.89 \mathrm{~m}
$$
