A force acts on a 3 ,g particle in such a way | vedclass

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Question

(A) According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.Giv

Vedclass · Accepted Answer

$528$

Vedclass · Answer

(A) According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.Given $x = 3t - 4t^2 + t^3$,the velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.At $t = 0 \,s$,$v_i = 3 - 8(0) + 3(0)^2 = 3 \,m/s$.At $t = 4 \,s$,$v_f = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \,m/s$.The mass $m = 3 \,g = 0.003 \,kg$.Substituting the values: $W = \frac{1}{2} 	imes 0.003 	imes (19^2 - 3^2) = 0.0015 	imes (361 - 9) = 0.0015 	imes 352 = 0.528 \,J$.Since $1 \,J = 1000 \,mJ$,the work done is $0.528 	imes 1000 = 528 \,mJ$.

$A$ force acts on a $3 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = 3t - 4t^2 + t^3$,where $x$ is in metres and $t$ is in seconds. The work done during the first $4 \,s$ is ..... $mJ$.

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