Question #108314

The brown hawk flies at an altitude of 25 meters at a speed of 6 m / s. The weight of the hawk is 600 g.


What is the a) kinetic energy of the hawk b) the potential energy and c) the mechanical energy when the zero level of the potential energy is at ground level?

Expert's answer

a) Kinetic energy of the hawk.

By definition, kinetic energy is: K=mv22K = \dfrac{mv^2}{2} , where mm is a mass, and vv is a velocity of body.

Thus, K=0.6622=10.8 J.K = \dfrac{0.6\cdot6^2}{2} = 10.8 \space\text{J}.


b) The potential energy.

By definition, the potential energy is: U=mghU = mgh, where mm is the mass of body, gg is the gravitational acceleration, hh is the height above the zero level of the potential energy.

Thus, U=0.69.825=147 JU = 0.6\cdot9.8\cdot25 = 147 \space\text{J}.


c) The mechanical energy.

By definition, the mechanical energy is the sum of kinetic and potential energies: W=T+UW = T+U .

Thus, W=10.8 J+147 J=157.8 JW = 10.8 \space\text{J} + 147\space\text{J} = 157.8\space\text{J} .


Answer. a) 10.8J, b) 147J, c) 157.8J


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