Question

A string is wrapped on a wheel of moment of inertia 0⋅20 kg-m2 and radius 10 cm and goes through a light pulley to support a block of mass 2⋅0 kg as shown in figure (10-E5). Find the acceleration of the block.
Figure

Answer 1

The acceleration of the block is 0.89 m/s².

Explanation:

Step 1:

           I = 0.20 kg-$m^2$  (Greatest pulley)

           r = 10 cm  

r converting centimeter to meter  :

           $r=\frac{10}{100} m=0.1 \mathrm{m}$ ( Lesser pulley is lighter )

Step 2:

The block's mass , m = 2 kg

Therefore,  

                mg – T = ma  -----> eqn ( 1 )

                         $T=\frac{l a}{r^{2}}$  -----> eqn ( 2 )

Equation (2) put in equation (1)

Step 3:

                   $\begin{aligned}&m g-\left(\frac{l a}{r^{2}}\right)=m a\\&m g=m a+\frac{l a}{r^{2}}\\&\mathrm{mg}=\left(\mathrm{m}+\frac{l}{r^{2}}\right) \mathrm{a}\\&a=\frac{m g}{\left(m+\frac{1}{r^{2}}\right)}\end{aligned}$      

g = Its gravity acceleration.  

Gravity acceleration value on Earth is 9.8 m / s².

Step 4:                          

Here, "a" is block’s acceleration.

                       $a=\frac{2 \times 9.8}{2+\frac{0.2}{0.01}}$

                       $a=\frac{19.6}{2+20}$

                       $a=\frac{19.6}{22}$

                      $a$  $=0.89 \mathrm{m} / s^{2}$

Hence, the block's acceleration = 0.89 m/$s^{2}$.                                                      

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

0.89 metre per second square