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

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The potential energy function (in J) of a particle in a region of space is given as U = (2x² + 3y³ + 2z). Here x, y and z are in meter. The magnitude of x - component of force (in N) acting on the particle at point P (1, 2, 3) m is:

Solution & Explanation

Related Formula

The force vector F is related to the potential energy U by the negative gradient of potential energy:

F = - ∇ U = -( (∂ U)/(∂ x) i + (∂ U)/(∂ y) j + (∂ U)/(∂ z) k )

Hence, the x-component of force is:

Fₓ = -(∂ U)/(∂ x)
Core Logic

Given the potential energy function:

U = 2x² + 3y³ + 2z

Taking the partial derivative with respect to x (treating y and z as constants):

(∂ U)/(∂ x) = (∂)/(∂ x)(2x²) = 4x
Step 1: Substitute Coordinates

The x-component of the force is:

Fₓ = -4x

At point P(1, 2, 3) ~m, we substitute x = 1:

Fₓ = -4(1) = -4 ~N

Magnitude of the x-component of force is:

|Fₓ| = 4 ~N
Pattern Recognition

When asked for a specific component (like x-component), only differentiate partially with respect to that specific variable. The remaining coordinates (y, z) act purely as constants and vanish.

Chapter Mix

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_2024_29_jan_morning

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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%)