The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x = 3 t^{2} + 4 t + 5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

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The displacement of a particle of mass 10g moving in one dimension under the action of a constant force 5N is related with time as t=√x+2, where x is in metre and 't' is in second. The work done by the force in first 5s will be.

The correct option is A 45JGiven, t=√x+2So, √x=t−2⇒x=t2−4t+4Also, F=5NSo, work done by this force in first 5s will beW=F⋅xt=5=5×[52−4×5+4]=5×[25−20+4]=5×9=45J.

The displacement of a particle of mass $10$ g moving in one dimension under the action of a constant force $5$ N is related with time as $t = \sqrt{x} + 2$ , where x is in metre and 't' is in second. The work done by the force in first $5$ s will be.

The correct option is A $45$ J
Given, $t = \sqrt{x} + 2$
So, $\sqrt{x} = t - 2$
$\Rightarrow x = t^{2} - 4 t + 4$
Also, $F = 5$ N
So, work done by this force in first $5$ s will be
$W = F \cdot x_{t} = 5$
$= 5 \times [5^{2} - 4 \times 5 + 4]$
$= 5 \times [25 - 20 + 4]$
$= 5 \times 9 = 45$ J.