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Byju's Answer

Standard XIII

Physics

Work Energy Theorem

The displacem...

Question

The displacement $x$ of a particle of mass $m kg$ moving in one dimension is related to the time $t$ by the equation $t = 4 x + 3$ where $x$ is in metre and $t$ is in second. The work done by the force in the first six seconds in joules is

0 J

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3 J

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6 J

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9 J

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Solution

The correct option is A $0 J$
Given,
$t = 4 x + 3 \Rightarrow x = \frac{1}{4} (t - 3)$
$v = \frac{d x}{d t} = \frac{1}{4} (1) = \frac{1}{4} m / s$
So, the velocity is constant.

According to Work energy theorem,
$W_{n e t} = Δ K E = 0$

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The displacement x of a particle of mass m kg moving in one dimension is related to the time t by the equation t=4x+3 where x is in metre and t is in second. The work done by the force in the first six seconds in joules is

The correct option is A 0 JGiven, t=4x+3⇒x=14(t−3) v=dxdt=14(1)=14 m/s So, the velocity is constant. According to Work energy theorem, Wnet=ΔKE=0

The displacement $x$ of a particle of mass $m kg$ moving in one dimension is related to the time $t$ by the equation $t = 4 x + 3$ where $x$ is in metre and $t$ is in second. The work done by the force in the first six seconds in joules is

The correct option is A $0 J$
Given,
$t = 4 x + 3 \Rightarrow x = \frac{1}{4} (t - 3)$
$v = \frac{d x}{d t} = \frac{1}{4} (1) = \frac{1}{4} m / s$
So, the velocity is constant.

According to Work energy theorem,
$W_{n e t} = Δ K E = 0$