JEE Main 2025 Apr 4 Shift 2, Physics Q14: Work Energy Theorem And Its Applications

Let the applied force be $F$ at $45^{\circ }$ above horizontal. Resolving:

Normal force balance: $N+\frac{F}{\sqrt{2}}=mg\Rightarrow N=mg-\frac{F}{\sqrt{2}}$.

The block moves with uniform velocity (the question implies sliding at constant speed), so the horizontal component of $F$ balances kinetic friction:

$\frac{F}{\sqrt{2}}=\mu N=\mu \left(mg-\frac{F}{\sqrt{2}}\right)$

$\frac{F}{\sqrt{2}}(1+\mu )=\mu mg$

$F=\frac{\mu mg\sqrt{2}}{1+\mu }=\frac{0.25\times 25\times 9.8\times \sqrt{2}}{1.25}=49\sqrt{2}\text{N}$.

Work done by $F$ over displacement $s=5$ m:

$W=F\cos 45^{\circ }\cdot s=49\sqrt{2}\cdot \frac{1}{\sqrt{2}}\cdot 5=49\times 5=245$ J.