JEE Main · Physics ↓ Falling

Work, Energy and Power appeared 31 times across 3 years — 3.6% of Physics. This question is from Work Done by Variable Force.

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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 t² + β t + γ m, where α = 1 m/s², 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.

Solution & Explanation

Related Formula
W = ∫ F · dx = F · Δ x (if force is constant) v = (dx)/(dt), a = (dv)/(dt)

F = m a

Core Logic

Given displacement equation:

x(t) = 1t² + 1t + 1

Velocity is the first derivative: v(t) = 2t + 1 Acceleration is the second derivative:

a(t) = 2 m/s²

Since acceleration is constant, the force is also constant:

F = m a = 2 × 2 = 4 N
Step 1: Finding Displacement Interval

We need the displacement during the interval t=2 to t=3:

x(3) = 3² + 3 + 1 = 9 + 3 + 1 = 13 m x(2) = 2² + 2 + 1 = 4 + 2 + 1 = 7 m

Total displacement S = x(3) - x(2) = 13 - 7 = 6 m

Step 2: Final Conclusion

Work done:

W = F · S = 4 N × 6 m = 24 J
Pattern Recognition

If x(t) is a quadratic in t, acceleration is constant. Work done simply becomes ma × Δ x.

Chapter Mix

Class 11 Physics: Kinematics Class 11 Physics: Work, Energy and Power

Reference Study Guides

Work Energy And Power

More Work, Energy and Power Previous-Year Questions — Page 7

Q49 jee_main_2024_31_jan_morning Conservation Of Momentum
An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :
  • A. M₁ / (M₁ + M₂)
  • B. (M₂)/(M₁)
  • C. M₂ / (M₁ + M₂)
  • D. (M₁)/(M₂)

Solution

Related Formula
KE = (p²)/(2m)
Core Logic

By conservation of linear momentum (since no external horizontal force acts on the system):

0 = M₁ v₁ + M₂ v₂ | p₁| = | p₂| = p

Both the artillery and the shell acquire the exact same magnitude of momentum during firing.

Step 2: Kinetic Energy Ratio

The kinetic energy is related to momentum by KE = (p²)/(2m). Since p is identical for both bodies:

KE ∝ (1)/(m)

Therefore, the ratio of kinetic energy of the artillery (M₁) to the shell (M₂) is:

KE₁KE₂ = ((p²)/(2M₁))/((p²)/(2M₂)) = (M₂)/(M₁)
Chapter Mix

Class 11 Physics: Work, Energy And Power

More Work, Energy and Power Questions — jee_main_2026_21_jan_evening

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Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)