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Standard X

Physics

Work

The displacem...

Question

The displacement $x$ in meter of a particle of mass $m k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $x = (t - 3)^{2}$ .The work done by the force (in joules) in first six seconds is

18 J

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Z e r o

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9 J

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36 J

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Solution

The correct option is C $Z e r o$
$x = {(t - 3)}^{2}$
Displacement after $6 s$ $,$ $x_{6} = {(6 - 3)}^{2} = 9$
Acceleration, $a = \frac{d^{2} x}{d t^{2}} = 2$
Mass $= m$
Work done, $W = m a x_{6}$
$= m \times 2 \times 9$
$= 18 m .$
Hence, the answer is $18 m .$

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The displacement x in meter of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in second by the equation x=(t−3)2.The work done by the force (in joules) in first six seconds is

The correct option is C Zerox=(t−3)2Displacement after 6s , x6=(6−3)2=9Acceleration, a=d2xdt2=2Mass =mWork done, W=max6 =m×2×9 =18m.Hence, the answer is 18m.

The displacement $x$ in meter of a particle of mass $m k g$ moving in one dimension under the action of a force is related to the time $t$ in second by the equation $x = (t - 3)^{2}$ .The work done by the force (in joules) in first six seconds is

The correct option is C $Z e r o$
$x = {(t - 3)}^{2}$
Displacement after $6 s$ $,$ $x_{6} = {(6 - 3)}^{2} = 9$
Acceleration, $a = \frac{d^{2} x}{d t^{2}} = 2$
Mass $= m$
Work done, $W = m a x_{6}$
$= m \times 2 \times 9$
$= 18 m .$
Hence, the answer is $18 m .$