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 ?

Question

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 ? (A) 850 J (B) 900 J (C) 950 J (D) 875 J

Tardigrade Admin · Accepted Answer

x = 3t2 + 5 v = (dx/dt) v = 6t +0 at t = 0 v = 0 t = 5 sec v = 30 m/s W.D. = ΔKE W.D. = (1/2 ) mv2 - 0 = (1/2)(2) (30)2 = 900 J

Q. 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

900 J

950 J

875 J

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

Q. A force acts on a 2kg 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 ?