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# A flying aeroplane has what kind of energy A. Only potential energyB. Only kinetic energy C. Both potential and kinetic energyD. None of these  Verified
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Hint: Every body or object has some energy associated with it. There are two kinds of energy associated with the motion of a body. If the body has some velocity then there will be some kinetic energy associated with it and if the body is at some height with respect to ground level then there will be some potential energy associated with it. So the total energy of the body will be the sum of its kinetic energy and potential energy.

As the aeroplane is flying it must have some velocity. So if the aeroplane has mass $M$and it is moving with velocity $v$ then its kinetic energy is given by
$K.E=\dfrac{1}{2}M{{v}^{2}}$.
Also the aeroplane moves at some height $H$with respect to the ground so it must have some potential energy, which is given by
$P.E=MgH$
So an aeroplane in motion has both kinetic energy and potential energy.
So the correct option is C.

Additional Information: The energy possessed by a body is measured in terms of capacity to do work. A moving object can do work. If the object is moving faster then it will do more work . The energy possessed by a body by virtue of its motion is called kinetic energy . If an object have initial velocity $u$and final velocity $v$ moving with uniform acceleration $a$ then the displacement is given by
\begin{align} & {{v}^{2}}-{{u}^{2}}=2as \\ & \Rightarrow s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a} \\ \end{align}
The force upon the body is given by$F=ma$
Work Done by the force$W=F.s=ma\times \dfrac{{{v}^{2}}-{{u}^{2}}}{2a}=\dfrac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)$
If initial velocity is zero i.e.$u=0$then $W=\dfrac{1}{2}m{{v}^{2}}$
This work done is a change in kinetic energy. so the kinetic energy possessed by a body having velocity $v$ is $\dfrac{1}{2}m{{v}^{2}}$.
Potential energy is the energy associated by a body by virtue of its position. If the object of mass $m$ is at height $h$ then the potential energy of the body is given by $P.E=mgh$.

Note: Note that total energy of any object will be equal to the sum of its kinetic energy and potential energy. if you throw a ball vertically upward then at start the potential energy is zero. As the height increases the potential energy increases and kinetic energy decreases. At the maximum height all of the kinetic energy is converted to potential energy and the kinetic energy is zero. When the object starts to fall from that height its kinetic energy will increase and potential energy decreases just before coming to the rest its kinetic energy is maximum. When the object comes to rest at ground its whole energy is transferred to the ground.