Question

The position x of particle moving along x-axis at time t given by equation t=rootx+2 where x is in meter and t is in seconds find work done by force in first four seconds

Answer 1

Given:

Position x of particle moving along x-axis at time t given by equation:

$\rm t = (\sqrt{x} + 2) \: s$

To Find:

Work done by force in first four seconds

Answer:

Position-time relation:

$\rm \sqrt{x} = t - 2 \\ \\ \rm x = ({(t - 2)}^{2} ) \: m \\ \\ \rm \dfrac{dx}{dt} = 2(t - 2) \\ \\ \rm dx = (2(t - 2)).dt \: \: \: \: \: \: \: \: \: ...eq_1$

Rate of change of position-time relation gives velocity:

$\rm v = \dfrac{dx}{dt} \\ \\ \rm v = \dfrac{d}{dt}( {(t - 2)}^{2} ) \\ \\ \rm v = 2(t - 2) \\ \\ \rm v = (2t - 4) \: m {s}^{ - 1}$

Rate of change of velocity-time relation gives accelration:

$\rm a = \dfrac{dv}{dt} \\ \\ \rm a = \dfrac{d}{dt}( 2t - 4) \\ \\ \rm a = 2 \: m {s}^{ - 1}$

As, acceleration of particle is constant hence force will be constant.

So,

$\rm \implies dW = F.dx \\ \\ \rm \implies \int dW = \int ma.dx \\ \\ \rm \implies W = ma \int dx \: \: \: \: \: \: ... from \ eq_1 \\ \\ \rm \implies W = ma \int\limits^4_0(2(t - 2)).dt \\ \\ \rm \implies W = m \times 2 \times 2 \int\limits^4_0(t - 2).dt \\ \\ \rm \implies W = 4m \int\limits^4_0(t - 2).dt \\ \\ \rm \implies W = 4m \bigg[ \dfrac{ {(t - 2)}^{2} }{2} \bigg]_0^4 \\ \\ \rm \implies W = 2m \bigg[ {(t - 2)}^{2} \bigg]_0^4 \\ \\ \rm \implies W = 2m \bigg[ {(4 - 2)}^{2} - {(0 - 2)}^{2} \bigg] \\ \\ \rm \implies W = 2m \bigg[ {2}^{2} - {( - 2)}^{2} \bigg] \\ \\ \rm \implies W = 2m \bigg[ 4 - 4 \bigg] \\ \\ \rm \implies W = 0$

$\therefore$ $\boxed{\mathfrak{Work \ done \ by \ force \ in \ first \ 4 \ s = 0}}$

Answer 2

Given:

$t = \sqrt{x} + 2$

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Need to find:

• The work done in first 4sec =?

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Solution:

$t = \sqrt{x} + 2 \\ \implies \: (t - 2) = \sqrt{x} \\ \implies \: x = {(t -2 )}^{2}$

We know,

$v = \dfrac{dx}{dt}$

$\implies \: v = \dfrac{d {(t - 2)}^{2} }{dx}$

$\implies \: v = \dfrac{d( {t}^{2} - 4t + 4) }{dx}$

$\purple{\bold{ \implies \: v = 2t - 4 \: m/s}}$

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Now we know,

$a = \dfrac{dv}{dt}$

$\implies \: a = \dfrac{d(2t - 4)}{dt}$

$\purple{\bold{\implies \: a = 2m \: / {s}^{2} }}$

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We know

$x = {(t - 2)}^{2}$

$\implies \: x = {t}^{2} - 4t + 4 \\ \implies \: \dfrac{dx}{dt} = \dfrac{d( {t}^{2} - 4t + 4) }{dt}$

$\purple{\bold{ \implies \: \dfrac{dx}{dt} = 2t - 4}}$

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dW= dF

$\implies \int dW = \int ma.dx$

As mass and acceleration is constant here,

$\implies W = ma \int dx$

$\implies W = ma \int\limits^4_0(2t - 4).dt$

taking 2 common,

$\implies W = m \times a \times 2 \int\limits^4_0(t - 2).dt$

$\implies W = 4m \int\limits^4_0(t - 2).dt$

$\implies W = 4m \bigg[ \dfrac{ {(t - 2)}^{2} }{2} \bigg]_0^4$

$\implies W = 2m \bigg[ {(t - 2)}^{2} \bigg]_0^4$

$\implies W = 2m \bigg[ {(4 - 2)}^{2} - {(0 - 2)}^{2} \bigg]$

$\implies W = 2m \bigg[ 4 - 4 \bigg]$

$\implies{\huge{\boxed{\red{\bold{W=0}}}}}$