JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A particle experiences a variable force \(\overrightarrow{ F }=\left(4 x \hat{ i }+3 y ^{2} \hat{ j }\right)\) in a horizontal \(x - y\) plane. Assume distance in meters and force is newton. If the particle moves from point \((1,2)\) to point \((2,3)\) in the \(x-y\) plane, the Kinetic Energy changes by............\(j\)
- A \(50.0\)
- B \(12.5\)
- C \(25.0\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(25.0\)
Step-by-step Solution
Detailed explanation
\(F =4 x \hat{ i }+3 y ^{2} \hat{ j }\) \(WD =\Delta KE\) \(W =\int \overrightarrow{ F } \cdot( dx \hat{ i }+ dy \hat{ j })\) \(=\int_{1}^{2} 4 xdx +\int_{2}^{3} 3 y ^{2} dx\) \(=\left(2 x ^{2}\right)_{1}^{2}+\left( y ^{3}\right)_{2}^{3}\) \(=(8-2)+(27-8)\) \(=6+19=25\, J\)
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