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Standard XII

Physics

Work Done When Force Is Varying

Under the act...

Question

Under the action of a force, a 2kg mass moves such that its position x as a function of time t is given by $x = t^{3} / 3$ where x is in meters and t in second. The work done by the force in first two seconds is:

1600 joule

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160 joule

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16 joule

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1.6 joule

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Solution

The correct option is C 16 joule
$x = \frac{t^{3}}{3}$
$a = \frac{d^{2} x}{d t^{2}} = 2 t$
Thus force acting on the particle= $m a = 2 m t$
Work done $= \int F d x$
$= \int_{0}^{2} (2 m t) t^{2} d t$
$= 2 m \int_{0}^{2} t^{3} d t$
$= 16 J$

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Under the action of a force, a 2kg mass moves such that its position x as a function of time t is given by x=t3/3 where x is in meters and t in second. The work done by the force in first two seconds is:

The correct option is C 16 joulex=t33a=d2xdt2=2tThus force acting on the particle=ma=2mtWork done=∫Fdx=∫20(2mt)t2dt=2m∫20t3dt=16J

Under the action of a force, a 2kg mass moves such that its position x as a function of time t is given by $x = t^{3} / 3$ where x is in meters and t in second. The work done by the force in first two seconds is:

The correct option is C 16 joule
$x = \frac{t^{3}}{3}$
$a = \frac{d^{2} x}{d t^{2}} = 2 t$
Thus force acting on the particle= $m a = 2 m t$
Work done $= \int F d x$
$= \int_{0}^{2} (2 m t) t^{2} d t$
$= 2 m \int_{0}^{2} t^{3} d t$
$= 16 J$