Logo
Jeeify
MCQEasyJEE 2024Friction (Static, Kinetic, Rolling)

JEE Physics 2024 Question with Solution

A block of mass 100 kg100 \, \text{kg} slides over a distance of 10 m10 \, \text{m} on a horizontal surface. If the coefficient of friction between the surfaces is 0.40.4, then the work done against friction (in J) is:

  • A

    4200 J4200 \, \text{J}

  • B

    3900 J3900 \, \text{J}

  • C

    4000 J4000 \, \text{J}

  • D

    4500 J4500 \, \text{J}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Mass of the block is 100 kg100 \, \text{kg}, distance is 10 m10 \, \text{m}, and coefficient of friction is 0.40.4 on a horizontal surface.

Find: The work done against friction.

The frictional force is

f=μNf = \mu N

On a horizontal surface, the normal reaction is

N=mgN = mg

Using the values shown in the solution,

f=μmg=0.4×100×10=400 Nf = \mu mg = 0.4 \times 100 \times 10 = 400 \, \text{N}

Now the work done against friction is

W=f×s=400×10=4000 JW = f \times s = 400 \times 10 = 4000 \, \text{J}

Therefore, the work done against friction is 4000 J4000 \, \text{J}, so the correct option is C. The first solution panel labels the option as D, but its own working concludes 4000 J4000 \, \text{J}, which matches option C.

Using friction and work formula

Given:

  • m=100 kgm = 100 \, \text{kg}
  • d=10 md = 10 \, \text{m}
  • μ=0.4\mu = 0.4

Find: Work done against friction.

Use the relation

W=f⋅dW = f \cdot d

where friction force is

f=μNf = \mu N

and for a horizontal surface

N=mgN = mg

Taking

g=9.8 m/s2g = 9.8 \, \text{m/s}^2

we get

N=100×9.8=980 NN = 100 \times 9.8 = 980 \, \text{N}

so

f=0.4×980=392 Nf = 0.4 \times 980 = 392 \, \text{N}

Hence,

W=392×10=3920 JW = 392 \times 10 = 3920 \, \text{J}

This is approximately

4000 J4000 \, \text{J}

which matches option C.

Therefore, from both the exact school-level approximation with g=10 m/s2g = 10 \, \text{m/s}^2 and the rounded value from g=9.8 m/s2g = 9.8 \, \text{m/s}^2, the defensible answer is C.

Common mistakes

  • Using the normal force incorrectly. On a horizontal surface, N=mgN = mg, not just mm. The friction formula is f=μNf = \mu N, so first compute the normal reaction properly.

  • Forgetting to multiply friction force by distance. Friction gives a force, but the question asks for work done. After finding ff, use W=fdW = fd.

  • Getting confused by the listed options label in the solution panel. The panel says option D, but the actual numerical working gives 4000 J4000 \, \text{J}, which corresponds to option C. Always trust the worked value over a mismatched label.

Practice more Friction (Static, Kinetic, Rolling) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Friction (Static, Kinetic, Rolling)

Related questions