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MCQEasyJEE 2024Work Done by Force

JEE Physics 2024 Question with Solution

A force (3x2+2x−5) N\left(3x^2 + 2x - 5\right) \, \text{N} displaces a body from x=2 mx = 2 \, \text{m} to x=4 mx = 4 \, \text{m}. The work done by this force is:

  • A

    50 J50 \, \text{J}

  • B

    58 J58 \, \text{J}

  • C

    62 J62 \, \text{J}

  • D

    70 J70 \, \text{J}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Force varies with position as F(x)=3x2+2x−5F(x) = 3x^2 + 2x - 5 and the body moves from x=2 mx = 2 \, \text{m} to x=4 mx = 4 \, \text{m}.

Find: The work done by the force.

For a variable force, work is the definite integral of force with respect to displacement:

W=∫24F(x) dx=∫24(3x2+2x−5) dxW = \int_{2}^{4} F(x) \, dx = \int_{2}^{4} \left(3x^2 + 2x - 5\right) \, dx

Now integrate term by term:

∫(3x2+2x−5) dx=x3+x2−5x+C\int \left(3x^2 + 2x - 5\right) \, dx = x^3 + x^2 - 5x + C

Apply the limits 22 to 44:

W=[x3+x2−5x]24W = \left[x^3 + x^2 - 5x\right]_{2}^{4} =(43+42−5×4)−(23+22−5×2)= \left(4^3 + 4^2 - 5 \times 4\right) - \left(2^3 + 2^2 - 5 \times 2\right)

Evaluate each bracket:

43=64,42=16,23=8,22=44^3 = 64, \quad 4^2 = 16, \quad 2^3 = 8, \quad 2^2 = 4

So,

W=(64+16−20)−(8+4−10)W = \left(64 + 16 - 20\right) - \left(8 + 4 - 10\right) =60−2=58 J= 60 - 2 = 58 \, \text{J}

Therefore, the work done is 58 J58 \, \text{J}. The correct option is B.

Average Force Check

Given: The displacement is from x=2 mx = 2 \, \text{m} to x=4 mx = 4 \, \text{m}.

Find: A quick consistency check for work.

The average force over the interval is

F‾=14−2∫24F(x) dx=582=29 N\overline{F} = \frac{1}{4-2}\int_{2}^{4} F(x) \, dx = \frac{58}{2} = 29 \, \text{N}

Then work can be checked as

W=F‾⋅Δx=29×2=58 JW = \overline{F} \cdot \Delta x = 29 \times 2 = 58 \, \text{J}

This confirms that the work done is 58 J58 \, \text{J}, so the correct option is B.

Common mistakes

  • Using F×sF \times s with a single force value is incorrect because the force changes with position. Instead, use the definite integral W=∫F(x) dxW = \int F(x) \, dx over the given limits.

  • Forgetting the lower limit contribution gives a wrong answer. After finding the antiderivative, always evaluate upper limit minus lower limit.

  • Making a sign error while integrating the constant term is common. Since ∫(−5) dx=−5x\int (-5) \, dx = -5x, the negative sign must be retained.

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