Question

Find $F$ such that the $2\text{kg}$ body goes up with an acceleration of $2{\text{m/s}}^2$.
Take ($g=10{\text{m/s}}^2$)

• A. $10\text{N}$
• B. $20\text{N}$
• C. $25\text{N}$
• D. $30\text{N}$

Answer 1

The correct option is D $30\text{N}$

FBDs of the two bodies are as shown.

For $2kg$ block, we can write
$T-2g=2a$
$\Rightarrow T=2(a+g)=2(2+10)=2\left(12\right)=24\text{N}$
For $3kg$ block, we can write
$F-T=3a$
$⟹F=T+3a$
$\therefore$ $F=24+3\left(2\right)=30$
Thus, $F=30\text{N}$

Alternate:
Consider the two bodies as a system whose acceleration is $2m/s^2$, so the effective force is on the left side.


So, for equilibrium we have
$a=\frac{\text{(Effective force)}}{\text{Total mass}}=\frac{F-2g}{5}$
$⟹2=\frac{F-2g}{5}⟹F=10+2g=10+2\times 10$
Thus, $F=30N$