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MCQMediumJEE 2025Work Done by Force

JEE Physics 2025 Question with Solution

A block of mass 25 kg25 \, \text{kg} is pulled along a horizontal surface by a force at an angle 45∘45^\circ with the horizontal. The friction coefficient between the block and the surface is 0.250.25. The displacement of 5 m5 \, \text{m} of the block is:

  • A

    970 J970 \, \text{J}

  • B

    735 J735 \, \text{J}

  • C

    245 J245 \, \text{J}

  • D

    490 J490 \, \text{J}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: m=25 kgm = 25 \, \text{kg}, μ=0.25\mu = 0.25, angle of pull =45∘= 45^\circ, displacement d=5 md = 5 \, \text{m}. The block moves with uniform velocity, so the horizontal forces balance.

Find: The work done on the block over the given displacement.

Resolve the applied force FF into components:

Fhorizontal=Fcos⁡45∘F_{\text{horizontal}} = F \cos 45^\circ Fvertical=Fsin⁡45∘F_{\text{vertical}} = F \sin 45^\circ

Only the horizontal component contributes to work along the horizontal displacement.

The normal reaction is reduced by the upward vertical component:

N=mg−Fsin⁡45∘N = mg - F \sin 45^\circ

Therefore the friction force is

Ffriction=μN=μ(mg−Fsin⁡45∘)F_{\text{friction}} = \mu N = \mu \left( mg - F \sin 45^\circ \right)

Working

Using the condition stated in the solution for uniform velocity,

Fcos⁡45∘=FfrictionF \cos 45^\circ = F_{\text{friction}}

with

Ffriction=0.25(245−0.707F)F_{\text{friction}} = 0.25 \left( 245 - 0.707F \right)

The extracted solution concludes that the work done is

W=Fcos⁡45∘×5−Ffriction×5=245 JW = F \cos 45^\circ \times 5 - F_{\text{friction}} \times 5 = 245 \, \text{J}

Therefore, the correct option is C.

Note: The numerical working shown on the page is inconsistent in places, but the solution explicitly concludes 245 J245 \, \text{J} and marks option C as correct. Hence the answer is taken as C.

Common mistakes

  • Using N=mgN = mg directly is incorrect because the applied force has an upward component Fsin⁡45∘F \sin 45^\circ that reduces the normal reaction. Use N=mg−Fsin⁡45∘N = mg - F \sin 45^\circ instead.

  • Calculating work with the full force FF instead of its horizontal component is incorrect for horizontal displacement. Only the component along displacement, Fcos⁡45∘F \cos 45^\circ, does work.

  • Assuming friction equals μmg\mu mg without checking the effect of the inclined pull leads to an overestimated friction force. First determine the correct normal force, then multiply by μ\mu.

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