A block of mass 2 kg is free to move along t | vedclass

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(C) The impulse imparted to the block is equal to the area under the $F-t$ graph.Impulse $J = \int F \ dt = 	ext{Area of triangle from } t=0 	ext{ t

Vedclass · Accepted Answer

$5.06$

Vedclass · Answer

(C) The impulse imparted to the block is equal to the area under the $F-t$ graph.Impulse $J = \int F \ dt = 	ext{Area of triangle from } t=0 	ext{ to } 3 \ s - 	ext{Area of triangle from } t=3 	ext{ to } 4.5 \ s$.At $t=0$,$F=4 \ N$. The slope of the line is $m = \frac{0-4}{3-0} = -\frac{4}{3} \ N/s$.So,$F(t) = 4 - \frac{4}{3}t$.At $t=4.5 \ s$,$F(4.5) = 4 - \frac{4}{3}(4.5) = 4 - 6 = -2 \ N$.Impulse $J = \left(\frac{1}{2} 	imes 3 	imes 4ight) - \left(\frac{1}{2} 	imes (4.5-3) 	imes 2ight) = 6 - 1.5 = 4.5 \ kg \cdot m/s$.Since the block starts from rest,the final momentum $p = J = 4.5 \ kg \cdot m/s$.The kinetic energy $K.E. = \frac{p^2}{2m} = \frac{(4.5)^2}{2 	imes 2} = \frac{20.25}{4} = 5.0625 \ J \approx 5.06 \ J$.

$A$ block of mass $2 \ kg$ is free to move along the $x$-axis. It is at rest and from $t=0$ onwards it is subjected to a time-dependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 \ s$ is (in $J$)

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