A force acts on a body of mass 10 kg, result | vedclass

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(D) Given, mass $m = 10 \ kg$ and displacement $x = \frac{t^3}{25} \ m$.Velocity $v = \frac{dx}{dt} = \frac{3t^2}{25} \ m/s$.Acceleration $a = \frac{d

Vedclass · Accepted Answer

$1.152$

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(D) Given, mass $m = 10 \ kg$ and displacement $x = \frac{t^3}{25} \ m$.Velocity $v = \frac{dx}{dt} = \frac{3t^2}{25} \ m/s$.Acceleration $a = \frac{dv}{dt} = \frac{6t}{25} \ m/s^2$.Force $F = m \cdot a = 10 \cdot \left(\frac{6t}{25}\right) = \frac{12t}{5} \ N$.Work done $dW = F \cdot dx = F \cdot \left(\frac{dx}{dt}\right) dt = \left(\frac{12t}{5}\right) \cdot \left(\frac{3t^2}{25}\right) dt = \frac{36t^3}{125} dt$.Total work done in the first $2 \ s$ is $W = \int_{0}^{2} \frac{36t^3}{125} dt = \frac{36}{125} \left[\frac{t^4}{4}\right]_{0}^{2} = \frac{36}{125} \cdot \frac{16}{4} = \frac{36 \cdot 4}{125} = \frac{144}{125} = 1.152 \ J$.

$A$ force acts on a body of mass $10 \ kg$, resulting in its displacement given as $x = \left(\frac{t^3}{25}\right) \ m$, where $t$ is the time in seconds. The work done by the force in the first $2 \ s$ is (in $J$)

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