Question

If $100N$ force applied to $10kg$ block as shown in the diagram, the acceleration of $40kg$ slab is

• A. $1.65m\cdot s^{-2}$
• B. $0.98m\cdot s^{-2}$
• C. $0.5m\cdot s^{-2}$
• D. $0.25m\cdot s^{-2}$

Answer 1

Answer: (B)

Static friction force between $10kg$ and $40kg$ block,

$F_s={\mu }_{s}R=0.6\times mg$
$=0.6\times 10\times 9.8=58.8N$
Here, we see that the applied force $(F=100N)$ is greater than friction force, hence $10kg$ block will start motion due to application of $100N$ force. Due to motion, kinetic friction force
$f_k={\mu }_{k}R=0.4mg$
$=0.4\times 10\times 9.8=39.2N$
$40kg$ body experiences a force of $f_k=39.2N$
$\therefore$ Acceleration of $40kg$ slab
$a=\frac{f_k}{40}=\frac{39.2}{40}=0.98m{s}^{-2}$