Under the action of a force, a 1 kg body moves, such that its position x as a function of time ‘t’ is given by x = t32, where x is in meter and t in second. The work done by the force in first 3 second is

x = t32 ⇒ ν = dxdt = 32 t2At t = o, νi = 0At t = 3sec, νf = 32 × 32 = 272 m/sw = Δk = Kf −Ki =12 mvt2 − 12 mvf2 w = 12 × 1 × (272)2 = 7298 J

Under the action of a force, a 1 kg body moves, such that its position x as a function of time ‘t’ is given by x = t32, where x is in meter and t in second. The work done by the force in first 3 second is

x = t32 ⇒ ν = dxdt = 32 t2At t = o, νi = 0At t = 3sec, νf = 32 × 32 = 272 m/sw = Δk = Kf −Ki =12 mvt2 − 12 mvf2 w = 12 × 1 × (272)2 = 7298 J

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Under the action of a force, a 1 kg body moves, such that its position $x$ as a function of time ‘t’ is given by $x = \frac{t^{3}}{2},$ where $x$ is in meter and t in second. The work done by the force in first 3 second is

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$\frac{243}{2} J$

$2430 J$

$\frac{729}{8} J$

24.35

answer is C.

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Detailed Solution

$x = \frac{t^{3}}{2} \Rightarrow ν = \frac{d x}{d t} = \frac{3}{2} t^{2}$
At $t = o, ν_{i} = 0$
At t = 3sec, $ν_{f} = \frac{3}{2} \times 3^{2} = \frac{27}{2} m/s$
$w = Δ k = K_{f} - K_{i} = \frac{1}{2} m v_{t}^{2} - \frac{1}{2} m v_{f}^{2}$

$w = \frac{1}{2} \times 1 \times {(\frac{27}{2})}^{2} = \frac{729}{8} J$