A force acts on a 2 kg object so that its position is given as a of time as x - 3t-2 - 5- What is the work done by this force in first 5 seconds -850 J950 J900 J875 J

Question

Toppr Admin · Accepted Answer

X-3t2-5v-dxdtv-6t-0att-0-xA0- -xA0-v-0t - 5 sec-xA0- -xA0-v - 30 m-sW-D- - -x394-KEW-D-12mv2-x2212-0-12-2-30-2-900J

A force acts on a $2 k g$ object so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first $5$ seconds ?
$850 J$
$950 J$
$900 J$
$875 J$

$x = 3 t^{2} + 5$
$v = \frac{d x}{d t}$
$v = 6 t + 0$
$a t t = 0$ $v = 0$
t = 5 sec v = 30 m/s
W.D. = $Δ K E$
$W . D . = \frac{1}{2} m v^{2} - 0 = \frac{1}{2} (2) (30)^{2} = 900 J$

A force acts on a 2 kg object so that its position is given as a function of time as x=3t2+5. What is the work done by this force in first 5 seconds ?850 J950 J900 J875 J

x=3t2+5v=dxdtv=6t+0att=0 v=0t = 5 sec v = 30 m/sW.D. = ΔKEW.D.=12mv2−0=12(2)(30)2=900J

A force acts on a $2 k g$ object so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first $5$ seconds ?
$850 J$
$950 J$
$900 J$
$875 J$

$x = 3 t^{2} + 5$
$v = \frac{d x}{d t}$
$v = 6 t + 0$
$a t t = 0$ $v = 0$
t = 5 sec v = 30 m/s
W.D. = $Δ K E$
$W . D . = \frac{1}{2} m v^{2} - 0 = \frac{1}{2} (2) (30)^{2} = 900 J$