A time-dependent force F = 6t acts on a parti | vedclass

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Question

(A) Given: Force $F = 6t$,mass $m = 1 \ kg$,initial velocity $u = 0$.Using Newton's second law,$F = ma = m \frac{dv}{dt}$.$6t = 1 \cdot \frac{dv}{dt}

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Given: Force $F = 6t$,mass $m = 1 \ kg$,initial velocity $u = 0$.Using Newton's second law,$F = ma = m \frac{dv}{dt}$.$6t = 1 \cdot \frac{dv}{dt} \implies dv = 6t \, dt$.Integrating from $t = 0$ to $t = 1 \ s$:$v = \int_{0}^{1} 6t \, dt = 6 \left[ \frac{t^2}{2} ight]_{0}^{1} = 3 \ m/s$.According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy:$W = \Delta KE = \frac{1}{2} m (v^2 - u^2)$.$W = \frac{1}{2} 	imes 1 	imes (3^2 - 0^2) = \frac{1}{2} 	imes 9 = 4.5 \ J$.

$A$ time-dependent force $F = 6t$ acts on a particle of mass $1 \ kg$. If the particle starts from rest,the work done by the force during the first $1 \ s$ will be ............... $J$.

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