JEE Main 2026 Jan 21 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

JEE Main2026Jan 21, Shift 2Physics

A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by $x (t) = α t^{2} + β t + γ$ m, where $α = 1$ m/s $^{2}$ , $β = 1$ m/s and $γ = 1$ m. The work done on the body during the time interval t = 2 s to t = 3 s, is _________ J.

A
49
B
42
C
24
D
12

Solution

With the given constants, $x (t) = t^{2} + t + 1$ . Velocity: $v (t) = 2 t + 1$ , acceleration $a = 2$ m/s $^{2}$ (constant).

Force: $F = ma = 2 \times 2 = 4$ N.

Displacement from $t = 2$ to $t = 3$ : $x (3) - x (2) = 13 - 7 = 6$ m.

Force is constant and along the motion, so $W = F \cdot s = 4 \times 6 = 24$ J.

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JEE Main 2026 Jan 21 Shift 2, Physics Q19: Work Energy Theorem And Its Applications

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