Answer:
135000 J
Step-by-step explanation:
Kinetic energy, KE is given by [tex]0.5mv^{2}[/tex] where m is the mass and v is the velocity.
Substituting 75 Kg for m and 60 m/s for v we obtain
[tex]KE=0.5*75*60^{2}=135000 J[/tex]
KE=135000J
The Kinetic energy of the considered club is evaluated to be [tex]16.875 \: \: \rm kg.m.s^{-1}[/tex]
Suppose the velocity of an object be [tex]v \: \rm m/s[/tex] and its mass be [tex]m \: \rm kg[/tex]
Then, its kinetic energy is given by:
[tex]KE = \dfrac{1}{2}\times m\times v^2 \: \: \rm kg.m.s^{-1}[/tex]
For this case, we have:
Therefore, the K.E. of the club is evaluated as:
[tex]KE = \dfrac{1}{2}\times m\times v^2 = \dfrac{60 \times 0.75^2}{2} = 16.875\: \: \rm kg.m.s^{-1}[/tex]
Thus, the Kinetic energy of the considered club is evaluated to be [tex]16.875 \: \: \rm kg.m.s^{-1}[/tex]
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