Question

A block of mass $10\text{kg}$ is put gently on a belt-conveyor system of infinite length at $t=0$, which is moving with constant speed $20\text{m/s}$ towards right at all time. A constant force of magnitude $15\text{N}$ is applied on the block continuously during its motion. Work done by the kinetic friction on the block of mass $10\text{kg}$ is

• A. $2500\text{J}$
• B. $-2500\text{J}$
• C. $1250\text{J}$
• D. $-1250\text{J}$

Answer 1

The correct option is C $1250\text{J}$

Here by the FBD of $10\text{kg}$ blockn conveyor belt will initially apply frictional force in the right direction
$\therefore$ kinetic friction = ${\mu }_{k}N=0.25\times 100$
$f_k=25N$
Applying $2^{nd}$ law of Newtons

$15+25=ma$

$\Rightarrow a=\frac{40}{10}=4{\text{m/s}}^2$
Now, distance travelled.
$S=\frac{v^2}{2a}=\frac{{20}^2}{2\times 4}=\frac{400}{8}=50\text{m}$

$\therefore$ Work done by kinetic friction on the block of mass $10\text{kg}=f_k.d$
$=f_k.d$
$=25\times 50$
$W_k=1250J$ Here the work done is positive as the frictional force and the displacement are in the same direction.

Hence option C is the correct answer