Logo
Jeeify
MCQEasyJEE 2025Work Done by Force

JEE Physics 2025 Question with Solution

A force F=α+βx2F = \alpha + \beta x^2 acts on an object in the xx-direction. The work done by the force is 5 J5 \, \text{J} when the object is displaced by 1 m1 \, \text{m}. If the constant α=1 N\alpha = 1 \, \text{N}, then β\beta will be:

  • A

    15 N/m215 \, \text{N/m}^2

  • B

    10 N/m210 \, \text{N/m}^2

  • C

    12 N/m212 \, \text{N/m}^2

  • D

    8 N/m28 \, \text{N/m}^2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: F=α+βx2F = \alpha + \beta x^2, α=1 N\alpha = 1 \, \text{N}, work done W=5 JW = 5 \, \text{J}, and displacement from x=0x = 0 to x=1 mx = 1 \, \text{m}.

Find: The value of β\beta.

For a variable force, work done is obtained by integrating force over displacement.

W=∫F dx=∫(α+βx2) dxW = \int F \, dx = \int (\alpha + \beta x^2) \, dx

Using the given values,

W=∫01(1+βx2) dx=[x+βx33]01W = \int_0^1 (1 + \beta x^2) \, dx = \left[ x + \frac{\beta x^3}{3} \right]_0^1

Substituting the limits,

W=(1+β3)−(0+0)=1+β3W = \left(1 + \frac{\beta}{3}\right) - (0 + 0) = 1 + \frac{\beta}{3}

Given that the work done is 5 J5 \, \text{J},

1+β3=51 + \frac{\beta}{3} = 5

Subtracting 11 from both sides,

β3=4\frac{\beta}{3} = 4

Multiplying by 33,

β=12\beta = 12

Therefore, the value of β\beta is 12 N/m212 \, \text{N/m}^2. The correct option is C.

Direct Integration Form

Given: F=α+βx2F = \alpha + \beta x^2 and W=5 JW = 5 \, \text{J} when x=1 mx = 1 \, \text{m}.

Find: The value of β\beta.

The work done by the variable force is written as

W=∫0xF dxW = \int_0^x F \, dx

Substituting F=α+βx2F = \alpha + \beta x^2,

W=∫0x(α+βx2) dxW = \int_0^x (\alpha + \beta x^2) \, dx

Evaluating the integral,

W=αx+βx33W = \alpha x + \frac{\beta x^3}{3}

Now put x=1 mx = 1 \, \text{m} and W=5 JW = 5 \, \text{J},

5=α⋅1+β⋅1335 = \alpha \cdot 1 + \frac{\beta \cdot 1^3}{3}

Since α=1 N\alpha = 1 \, \text{N},

5=1+β35 = 1 + \frac{\beta}{3}

Hence,

5−1=β35 - 1 = \frac{\beta}{3}4=β34 = \frac{\beta}{3}β=12 N/m2\beta = 12 \, \text{N/m}^2

Therefore, the correct value is 12 N/m212 \, \text{N/m}^2, so the correct option is C.

Common mistakes

  • Using W=F⋅xW = F \cdot x directly with a single constant force value is incorrect because the force depends on xx. For a variable force, evaluate work using the integral W=∫F dxW = \int F \, dx instead.

  • Forgetting the limits of integration from x=0x = 0 to x=1 mx = 1 \, \text{m} gives an incomplete expression for work. Use the definite integral with the stated displacement range.

  • Integrating βx2\beta x^2 incorrectly as βx2/2\beta x^2/2 is a conceptual error. The correct antiderivative is βx33\frac{\beta x^3}{3}.

Practice more Work Done by Force questions

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

Start free trialMore on Work Done by Force

Related questions