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

Question

Dr. Abubacker · Accepted Answer

(d) $x = 2t^{4} + 5t + 4 = v = \frac{dx}{dt} = 8t + 5$
At $t = 0,v = 5\ m/s$
At $t = 1s,\ v = 8 	imes 1 + 5 = 13\ m/s$
Increase in KE
$= \frac{1}{2}m\left\lbrack \left( 13^{\ } ight)^{2} - (5)^{2} ightbrack = 144\ J$

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

Explanation

Reviewed by Dr. Abubacker

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