
A hawk of mass \[10kg\] dives to catch a mouse. During dive, it loses \[500J\] of its potential energy. The height from where it begins its dive is? (Take \[g = 10{\text{ }}m/{s^2}\]
Answer
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Hint: Gravitational potential energy is said to be the energy possessed or acquired by an object due to a variation in its position when it is present in a gravitational field. In normal terms, it can be understood that gravitational potential energy is the energy of an object that is connected to gravitational force or gravity. But the equation \[\Delta P{E_g}\; = \;mgh,\] is represented to indicate the change in its gravitational potential energy.
Complete step by step solution:
We can say that the potential energy of a body at a given position is defined as the energy stored in the body at that position. When the position of the body varies due to the application of external forces the change in potential energy is identical to the amount of work done on the body by the forces.
The change in gravitational potential energy \[\Delta P{E_g}\] is \[\Delta P{E_g}\; = \;mgh,\] where $h$= the increase in height and $g$ is the acceleration due to gravity. The gravitational potential energy of a body near the surface of the earth is due to its place in the mass-Earth system. The only change is gravitational potential energy, (\[\Delta P{E_g}\]). As an object comes down without friction, its gravitational potential energy changes into kinetic energy equivalent to increasing speed, so that \[\Delta KE\]= \[ - \Delta P{E_g}\]
So we can say that
Change in Gravitational Potential Energy = mass $\times $acceleration $\times$ change in height
\[500{\text{ }}J = 10{\text{ }}kg \times 10 \times {\text{change in height}}\]
Change in height = \[5m\]
So, the hawk will dive to catch the mouse at a distance of \[5m\]
Note:
Because of gravitational force, the work done is not dependent on the path taken for a change in position so the force can be called conservative force. Also, all such forces have some potential in them. The gravitational effect on a body at infinity is equal to zero. Therefore, potential energy is zero, it is known as a reference point. Work done opposing gravity in lifting an object converts to the potential energy of the object-Earth system.
Complete step by step solution:
We can say that the potential energy of a body at a given position is defined as the energy stored in the body at that position. When the position of the body varies due to the application of external forces the change in potential energy is identical to the amount of work done on the body by the forces.
The change in gravitational potential energy \[\Delta P{E_g}\] is \[\Delta P{E_g}\; = \;mgh,\] where $h$= the increase in height and $g$ is the acceleration due to gravity. The gravitational potential energy of a body near the surface of the earth is due to its place in the mass-Earth system. The only change is gravitational potential energy, (\[\Delta P{E_g}\]). As an object comes down without friction, its gravitational potential energy changes into kinetic energy equivalent to increasing speed, so that \[\Delta KE\]= \[ - \Delta P{E_g}\]
So we can say that
Change in Gravitational Potential Energy = mass $\times $acceleration $\times$ change in height
\[500{\text{ }}J = 10{\text{ }}kg \times 10 \times {\text{change in height}}\]
Change in height = \[5m\]
So, the hawk will dive to catch the mouse at a distance of \[5m\]
Note:
Because of gravitational force, the work done is not dependent on the path taken for a change in position so the force can be called conservative force. Also, all such forces have some potential in them. The gravitational effect on a body at infinity is equal to zero. Therefore, potential energy is zero, it is known as a reference point. Work done opposing gravity in lifting an object converts to the potential energy of the object-Earth system.
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