Question

The vertical displacement of block $\mathrm{A}$in the meter is given by $y=\frac{t^2}{4}$, where t is in second. Calculate the downward acceleration of block $\mathrm{B}$.

Answer 1

Step 1: Given data

Vertical displacement of block A in the meter is given by $y=\frac{t^2}{4}$where$y$is the displacement and $t$ is the parameter on which $y$ depends.

Step 2: Formula Used

Acceleration $a=\frac{d{v}^{}}{dt}=\frac{d^{2}y}{d{t}^2}$

where $a,v,y,$ and $t$ are acceleration, velocity, displacement, and time respectively.

Step 3: Calculating the acceleration on B

As block A moves by $y$, the pulley above it will displace by $2y$, same goes with the pulley system that holds block $\mathrm{B}$, which will be displaced by $8y$ as shown in the figure above.

Thus, if block $\mathrm{A}$ moves by $y$, then the displacement of $\mathrm{B}$ will be $8y$.

Where $S_{A}and{S}_B$ are displacement of $\mathrm{A}$ and $\mathrm{B}$ respectively.

Now,

$$\begin{gathered} \frac{dy}{dt}=\frac{2t}{4}=\frac{t}{2} \\ \frac{d^{2}y}{d{t}^2}=\frac{1}{2} \end{gathered}$$

Now, acceleration on $\mathrm{B}$,

$$\begin{gathered} a_B=\frac{d{v}_B}{dt} \\ and{v}_B=\frac{d{S}_B}{dt}=8\frac{dy}{dt} \\ \Rightarrow a_B=\frac{d{v}_B}{dt}=8\frac{d^{2}y}{d{t}^2} \\ \Rightarrow a_B=8\times \frac{1}{2}=4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, acceleration on block B is $4\mathrm{m}/{\mathrm{s}}^2$.