A force acts on a 30 ,g particle in such a wa | vedclass

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(A) Given, mass of particle, $m = 30 \,g = 3 	imes 10^{-2} \,kg$.The position of the particle is $x = \alpha t^2$.Velocity $v = \frac{dx}{dt} = 2\alp

Vedclass · Accepted Answer

$0.96$

Vedclass · Answer

(A) Given, mass of particle, $m = 30 \,g = 3 	imes 10^{-2} \,kg$.The position of the particle is $x = \alpha t^2$.Velocity $v = \frac{dx}{dt} = 2\alpha t$.Acceleration $a = \frac{dv}{dt} = 2\alpha$.Force $F = ma = (3 	imes 10^{-2} \,kg) 	imes (2 	imes 1 \,m/s^2) = 6 	imes 10^{-2} \,N$.Work done $W = \int F dx$. Since $x = \alpha t^2$, $dx = 2\alpha t dt$.At $t = 0, x = 0$. At $t = 4, x = 1 	imes (4)^2 = 16 \,m$.$W = \int_{0}^{16} (6 	imes 10^{-2}) dx = 6 	imes 10^{-2} 	imes [x]_{0}^{16} = 6 	imes 10^{-2} 	imes 16 = 0.96 \,J$.

$A$ force acts on a $30 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = \alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

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