Question

An object of mass 10 kg is moving with a velocity of 20m/s. A force of 60N acted upon it for 2.3s. what will be the percentage increase in it kinetic energy​

Answer 1

Answer:

An object of mass 10 kg is moving with a velocity of 20m/s. A force of 60N acted upon it for 2.3s, the percentage increase in kinetic energy​ is 186%

Explanation:

The mass of the given body $m=10kg$

The initial velocity of the body $u=20m/s$

so that the initial kinetic energy of the body $KE_{i} =\frac{1}{2} mu^{2}$

⇒ $KE_{i} =\frac{1}{2} *10*20^{2}=2000J$

The force acted upon the body $F=60N$

duration of application of the force $t=2.3 s$

using Newton's law we have $F=ma$

substituting the given values we get the acceleration of the body $a=F/m$

⇒$a=60/10=6m/s^{2}$

now using the equation of motion $V=u+at$

we get the final velocity of the body $V=20+6*2.3=33.8m/s$

new kinetic energy $KE_{f}=\frac{1}{2} mV^{2}$

⇒ $KE_{f}=\frac{1}{2} *10*33.8^{2}=5712.2J$

the increase in the kinetic energy $5712.2-2000=3712.2J$

therefore the percentage increase in kinetic energy

$\frac{3712.2}{2000} *100=185.61$ %