A particle experiences a variable force $\overrightarrow{\mathrm{F}}=\left(4x\hat{\mathrm{i}}+3{\mathrm{y}}^2\hat{\mathrm{j}}\right)$ in a horizontal x-y plane. Assume distance in meters and force in newton. If the particle moves from point (1,2) to point (2,3) in the x-y plane, the Kinetic Energy changes by

A particle experiences a variable force f=4xi in a horizontal x-y plane. The particle is influenced by a force expressed as f=(4xi + 3y²j), which varies with position. To determine the change in kinetic energy as the particle moves from point (1, 2) to point (2, 3), we can use the work-energy theorem.

According to the work-energy theorem, the work done by the variable force is equal to the change in kinetic energy. Since a particle experiences a variable force f=4xi+3y²j, we can calculate the work done using the integral:

W = ∫ F ⋅ (dx i + dy j)

Substituting the given force f=(4xi + 3y²j) into the equation:

W = ∫₁² 4x dx + ∫₂³ 3y² dy

Solving each integral separately:

∫₁² 4x dx = [ 2x² ]₁² = (2 × 2²) - (2 × 1²) = 8 - 2 = 6

∫₂³ 3y² dy = [ y³ ]₂³ = (3³) - (2³) = 27 - 8 = 19

Adding both values, the total work done by the variable force f=4xi+3y²j is:

W = 6 + 19 = 25 J

Therefore, under the action of a variable force f 4xi 2yj, the kinetic energy of the particle changes by 25 J as it moves from point (1, 2) to point (2, 3).

This calculation shows how a particle experiences a variable force f=4xi in combination with a force term 3y²j, leading to a total kinetic energy change of 25 J. Understanding how a force f=(4xi + 3yj) affects the particle's motion is crucial in analyzing motion under varying conditions.