Given that the position of the body in m is a function of time as follows: x = 2t4 + 5t + 4 The mass of the body is 2 kg. What is the increase in its kinetic energy one second after the start of motion?

Question

Given that the position of the body in m is a function of time as follows: x = 2t4 + 5t + 4 The mass of the body is 2 kg. What is the increase in its kinetic energy one second after the start of motion? (A) 168J (B) 169J (C) 32 J (D) 144 J

Tardigrade Admin · Accepted Answer

x=2 t4+5 t+4 ⇒ v=(d x/d t)=8 t+5 at t=0, v=5 m / s at t=1 s , v=8 × 1+5=13 m / s Increases in KE =(1/2) m [132-52] =144 J

Q. Given that the position of the body in m is a function of time as follows: $x = 2 t^{4} + 5 t + 4$ The mass of the body is $2 k g$ . What is the increase in its kinetic energy one second after the start of motion?

168J

169J

32 J

144 J

$x = 2 t^{4} + 5 t + 4$
$\Rightarrow v = \frac{d x}{d t} = 8 t + 5$
at $t = 0, v = 5 m / s$
at $t = 1 s, v = 8 \times 1 + 5 = 13 m / s$
Increases in $K E = \frac{1}{2} m [1 3^{2} - 5^{2}]$
$= 144 J$

Q. Given that the position of the body in m is a function of time as follows: x=2t4+5t+4 The mass of the body is 2kg. What is the increase in its kinetic energy one second after the start of motion?