Physics Q44 – JEE-MAIN 2025

Home Physics Work, Energy & PowerQuestion #20

A block of mass

25 kg

is pulled along a horizontal surface by a force at an angle

45^{\circ}

with the horizontal. The friction coefficient between the block and the surface is

0.25

. The block travels at a uniform velocity. The workdone by the applied force during a displacement of

5 m

of the block is :

Clue 1 of 3: Equilibrium Condition

Since the block travels at a uniform velocity, the net force acting on it is zero.

🥷

Ninja Strategy•Maximum Possible Work Check

Calculate the maximum possible work done against friction (assuming a purely horizontal applied force, which maximizes normal force and thus friction). The actual work done must be less than this value, allowing elimination of options that are too large.

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Step 1: Analyze Forces and Equilibrium✦ Active

Let the applied force be $F$ . Its horizontal component is $F_{x} = F \cos 45^{\circ}$ and its vertical component is $F_{y} = F \sin 45^{\circ}$ . Since the block moves at a uniform velocity, the net force in both horizontal and vertical directions is zero.

Vertical equilibrium: The normal force $N$ acts upwards, gravity $m g$ acts downwards, and $F_{y}$ acts upwards. So, $N + F \sin 45^{\circ} = m g$ . This gives $N = m g - F \sin 45^{\circ}$ .

Horizontal equilibrium: The horizontal component of the applied force $F_{x}$ acts in the direction of motion, and the kinetic friction force $f_{k}$ acts opposite to it. So, $F \cos 45^{\circ} = f_{k}$ .

The kinetic friction force is $f_{k} = μ N$ . Substituting $N$ : $f_{k} = μ (m g - F \sin 45^{\circ})$ .

Step 2: Calculate Applied Force○ Expand

Equating the horizontal forces: $F \cos 45^{\circ} = μ (m g - F \sin 45^{\circ})$ .

Rearranging to solve for $F$ : $F \cos 45^{\circ} + μ F \sin 45^{\circ} = μ m g ⟹ F (\cos 45^{\circ} + μ \sin 45^{\circ}) = μ m g$ .

Given $m = 25 kg$ , $μ = 0.25$ , $θ = 45^{\circ}$ , $g = 9.8 {m/s}^{2}$ . Also, $\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$ .

F (\frac{1}{\sqrt{2}} + 0.25 \frac{1}{\sqrt{2}}) = 0.25 \times 25 \times 9.8

F \frac{1}{\sqrt{2}} (1 + 0.25) = 0.25 \times 25 \times 9.8

F = \frac{0.25 \times 25 \times 9.8 \times \sqrt{2}}{1.25} = \frac{61.25 \times 1.414}{1.25} \approx 69.3 N

💡 Teacher's Secret Hint

Remember to use $g = 9.8 {m/s}^{2}$ for precise calculations unless specified otherwise.

Step 3: Calculate Work Done by Applied Force○ Expand

The work done by the applied force $W_{F}$ during a displacement $d$ is given by $W_{F} = F d \cos θ$ .

Using the expression for $F$ from Step 2: $W_{F} = (\frac{μ m g \sqrt{2}}{1.25}) d \cos 45^{\circ}$ .

Substitute $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ :

W_{F} = \frac{μ m g \sqrt{2}}{1.25} \times d \times \frac{1}{\sqrt{2}} = \frac{μ m g d}{1.25}

Now, substitute the given values: $μ = 0.25$ , $m = 25 kg$ , $g = 9.8 {m/s}^{2}$ , $d = 5 m$ .

W_{F} = \frac{0.25 \times 25 \times 9.8 \times 5}{1.25}

Since $1.25 = 5 \times 0.25$ , the expression simplifies to:

W_{F} = \frac{0.25 \times 25 \times 9.8 \times 5}{5 \times 0.25} = 25 \times 9.8 = 245 J

💡 Teacher's Secret Hint

Alternatively, since the block moves at uniform velocity, the work done by the applied force is equal to the work done against friction, $W_{F} = f_{k} d$ . And $f_{k} = F \cos 45^{\circ}$ . So $W_{F} = (F \cos 45^{\circ}) d$ . The simplified formula $W_{F} = \frac{μ m g d}{1 + μ}$ can also be used directly.

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Physics Question 44 – JEE-MAIN 2025

Clue 1 of 3: Equilibrium Condition

🧠 Full Solution Path