Question

An inclined plane making an angle of 25 degrees with the horizontal has a pulley at its top. A 30-kg block on the plane is connected to a freely hanging 20-kg block by means of a cord passing over the pulley. Compute the distance the 20-kg block will fall in 2 s starting from rest. Neglect friction.

Answer 1

Given:

Let

$\begin{array}{l}\mathrm{M}_{A}=30 \mathrm{~kg} \\\mathrm{M}_{B}=20 \mathrm{~kg}\end{array}$

For block B

$\begin{array}{l}M_{B} g-T=M_{b} a \\20 \times 9.8-T=20 a \\T+20 a=196 - (1)\end{array}$

For block A

$\begin{array}{l}T-M_{A} g \sin 25=M_{a} a \\T-30 \times 9.8 \times \sin 25=30 a \mathrm{a} \\T-30 a=124.25 - (2)\end{array}$

From 1 and 2

$\begin{array}{l}50 a=196-124.25 \\a=1.435 \mathrm{~m} / \mathrm{s}^{2}\end{array}$

$\text { So distance traveled by } 20 \mathrm{~kg} \text { block is }$

$\begin{array}{l}Y=V_{o y} t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 1.435 \times 2^{2} \\Y=2.87 \mathrm{~m}\end{array}$