Question

The $100kg$block in FIGURE EX$7.25$takes $6.0s$to reach the floor after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.

Short answer

The acceleration of the$2.0kg$blocks is$2.16m/s^2$

Step-by-step solution

Given information 

Mass of right block, $M=100kg$

Time taken $t=6s$

Distance moved$s=1m$

Explanation

Equation of motion for mass m

$T-mg=ma\dots \dots \left(1\right)$

Equation of motion for mass M:

$Mg-T=Ma\cdots \cdots \left(2\right)$

On adding equation $\left(1\right)\text{and(2)}$

$$\begin{gathered} g\left(M-m\right)=a\left(M+m\right) \\ m=\frac{M\left(g-a\right)}{\left(g+a\right)} \end{gathered}$$

Acceleration a can be calculated as follows.

localid="1649613906426" $$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ u=0 \\ s=1m \\ t=6s \\ 1m=0+\frac{1}{2}\left(a\right){\left(6s\right)}^2 \\ a=\frac{2}{36}m/s^2 \\ a=0.055m/s^2 \\ m=\frac{\left(100kg\right)\left(9.8m/s^2-0.055m/s^2\right)}{\left(9.8m/s^2\right)+\left(0.055m/s^2\right)} \\ m=98.87kg \end{gathered}$$

The mass of the block on the left side is$98.87kg$