Question

Consider the situation as shown in figure. A constant force acting on the lower blocks 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. $8\text{J}$
• B. Zero
• C. $16\text{J}$
• D. $10\text{J}$

Answer 1

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

By the $FBD$ 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 in $2\text{kg}$ body

$\therefore$ By 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 $=f_s.S$
As displacement is given as $(S=2\text{m})$
$W=8\times 2$
$W=16\text{J}$

Hence option C is the correct answer