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"rocket man" has a propulsion unit strapped to his back. he starts from rest on the ground, fires the unit, and accelerates straight upward. at a height of 15 m, his speed is 4.8 m/s. his mass, including the propulsion unit, has the approximately constant value of 135 kg. find the work done by the force generated by the propulsion unit. (neglect air resistance.)

Respuesta :

skyluke89
For the work-energy theorem, the work done by the propulsion unit should be equal to the total mechanical energy E gained by the rocket man:
[tex]W=E=K+U[/tex]
where K is the kinetic energy gained by the rocket man, while U is its gravitational potential energy at height h=15 m, where his speed is v=4.8 m/s. Expliciting the terms, we find the work:
[tex]W=K+U= \frac{1}{2}mv^2+mgh=[/tex]
[tex]= \frac{1}{2}(135 kg)(4.8 m/s)^2+(135 kg)(9.81 m/s^2)( 15 m)=21420 J [/tex]

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