A force F = ( 3 t i ^ + 5 j ^ ) N acts on a body due to which its displacement varies as s = ( 2 t 2 i ^ − 5 j ^ ) . Work done by this force in t = 0 to 2 sec is:

F = 3 t i ^ + 5 j ^ S = 2 t 2 i ^ − 5 j ^ dt d s = 4 t i ^ ω = ∫ F ⋅ d s = ∫ 0 2 ( 3 t i ^ + 5 j ^ ) ⋅ ( 4 t i ^ ) dt = ∫ 0 2 12 t 2 dt = 4 ( 2 3 − 0 2 ) = 32 J

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PhysicsW.E.P.Calculation of workMedium2 minQB

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A force $F = (3 t \hat{i} + 5 \hat{j}) N$ acts on a body due to which its displacement varies as $s = (2 t^{2} \hat{i} - 5 \hat{j})$ . Work done by this force in $t = 0$ to 2 sec is:

Options:

Answer:

Solution:

$F = 3 t \hat{i} + 5 \hat{j}$

$S = 2 t^{2} \hat{i} - 5 \hat{j} \frac{d s}{dt} = 4 t \hat{i} ω = \int F \cdot d s = \int_{0}^{2} (3 t \hat{i} + 5 \hat{j}) \cdot (4 t \hat{i}) dt = \int_{0}^{2} 12 t^{2} dt = 4 (2^{3} - 0^{2}) = 32 J$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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