A bead slides along a frictionless wire lying on a horizontal plane from origin to point P. Wire makes the angle of 45° with x-axis as shown in figure. In addition to any normal forces exerted by the wire, the bead is subject to an external force that depends on position according to formula F = F₀(x/x₀)²î+F₀(y/y₀)²ĵ. Find the work done by the force F (vector) (in J) on bead till it reaches at end P of the wire. (Given data: F₀=1 N, x₀=6 m, mass of bead=2 kg)
Physics · Mixed Chapter
JEE Mains

Question

A bead slides along a frictionless wire lying on a horizontal plane from origin to point P. Wire makes the angle of 45° with x-axis as shown in figure. In addition to any normal forces exerted by the wire, the bead is subject to an external force that depends on position according to formula F = F₀(x/x₀)²î+F₀(y/y₀)²ĵ. Find the work done by the force F (vector) (in J) on bead till it reaches at end P of the wire. (Given data: F₀=1 N, x₀=6 m, mass of bead=2 kg)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The bead follows the 45° line from (0,0) to P=(x₀,y₀), where y₀=x₀ because tan45°=1. Step 2: The work by the position-dependent external force is W=∫F·dr=∫₀ˣ⁰F₀(x/x₀)²dx +∫₀ʸ⁰F₀(y/y₀)²dy. Step 3: Each integral equals F₀ times its endpoint distance divided by 3. Thus W=(F₀/3)(x₀+y₀)=2F₀x₀/3. Step 4: W=2×1×6/3=4 J.