A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by x = 3t - 4t² + t³, where x is in metre and t in second. The work done during the first 4 s is (a) 570 mJ (b) 450 mJ (c) 490 mJ (d) 528 mJ

Answer: (d). v = dx/dt = 3 - 8t + 3t² v(0) = 3 m/s, v(4) = 3 - 32 + 48 = 19 m/s W = Δ K = 1/2(3 × 10⁻³)(19² - 3²) = 1.5 × 10⁻³ × 352 = 0.528 J So the answer is (d) 528 mJ.

A force acts on a 3.0 g particle in such a way that the position of… | NEET 1998 (AIPMT)

Q30NEET 1998 · AIPMTMedium

Work-Energy Theorem

A force acts on a $3.0\text{ g}$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$ , where $x$ is in metre and $t$ in second. The work done during the first $4\text{ s}$ is

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Answer: (d)

$v = \frac{dx}{dt} = 3 - 8t + 3t^2$
$v(0) = 3\text{ m/s}$ , $v(4) = 3 - 32 + 48 = 19\text{ m/s}$
$W = \Delta K = \frac{1}{2}(3 \times 10^{-3})(19^2 - 3^2) = 1.5 \times 10^{-3} \times 352 = 0.528\text{ J}$
So the answer is (d) $528\text{ mJ}$ .

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