Answer:
a) [tex]W_{F} = 17278.144\,J[/tex], b) [tex]W_{g} = 15838.305\,J[/tex], c) [tex]K = 1440.071\,J[/tex], d) [tex]v \approx 5.821\,\frac{m}{s}[/tex]
Explanation:
a) The equation of equilibrium for the astronaut is:
[tex]\Sigma F_{y} = F - m\cdot g = \frac{m\cdot g}{11}[/tex]
The expression for the force from the helicopter is:
[tex]F = \frac{12\cdot m \cdot g}{11}[/tex]
[tex]F = \frac{12\cdot(85\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{11}[/tex]
[tex]F = 909.376\,N[/tex]
The work done by the force is:
[tex]W_{F} = (909.376\,N)\cdot (19\,m)[/tex]
[tex]W_{F} = 17278.144\,J[/tex]
b) The work done by the gravitational force on the astronaut is:
[tex]W_{g} = (85\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m)[/tex]
[tex]W_{g} = 15838.305\,J[/tex]
d) The speed before the astronaut reaches the helicopter is:
[tex]v = \sqrt{2\cdot \frac{g}{11}\cdot s }[/tex]
[tex]v =\sqrt{\frac{2\cdot g\cdot s}{11} }[/tex]
[tex]v=\sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m)}{11} }[/tex]
[tex]v \approx 5.821\,\frac{m}{s}[/tex]
c) The kinetic energy of the astronaut before reaching the helicopter is:
[tex]K = \frac{1}{2}\cdot (85\,kg)\cdot (5.821\,\frac{m}{s} )^{2}[/tex]
[tex]K = 1440.071\,J[/tex]
Which is equal to the net work on the astronaut.
