Question

Consider the situation as shown in the figure. A constant force acting on the lower block produces an acceleration of $4{\text{m/s}}^2.$ The two blocks always move together. Find the work done by static friction on the upper block during displacement $2\text{m}$ of the system.

• A. Zero
• B. $16\text{J}$
• C. $10\text{J}$
• D. $8\text{J}$

Answer 1

The correct option is B $16\text{J}$

By the $\text{FBD}$ of $2\text{kg}$ body, there will be only frictional force in the forward direction for the block to move forward
with acceleration, $a=4{\text{m/s}}^2$


Now since the whole system is moving at $a=4{\text{m/s}}^2$ and $f_s$ is the only force acting on $2\text{kg}$ body.

From Newton's $2^{nd}$ law,

$f_s=ma=2\times 4=8\text{N}$
(As both blocks move together)

$\therefore$ Work done by static friction on upper block,

$W=f_s\cdot S$
$W=8\times 2(\text{Given}:S=2\text{m})$
$W=16\text{J}$

Hence, option $\left(\text{C}\right)$ is the correct answer.