Respuesta :
(a) -1620.8 J
The initial kinetic energy of the skateboarder is:
[tex]K_i = \frac{1}{2}mu^2 = \frac{1}{2}(55.6 kg)(2.44 m/s)^2=165.5 J[/tex]
where m is the skateboarder's mass and u his initial speed;
While the final kinetic energy is
[tex]K_f = \frac{1}{2}mv^2 = \frac{1}{2}(55.6 kg)(7.24 m/s)^2=1457.2 J[/tex]
where v is his final speed.
So the change in kinetic energy is
[tex]\Delta K=K_f - K_i = 1457.2 J -165.6 J = 1291.6 J[/tex]
According to the work-energy theorem, the change in mechanical energy (kinetic+potential) of the skateboarder is equal to the work done on it:
[tex]\Delta K + \Delta E_p = W + W_f[/tex]
where
[tex]W = 80.4 J[/tex] is the work done by the skateboarder on himself
[tex]W_f = -244 J[/tex] is the work done by friction
[tex]\Delta E_p = E_p_f - E_p_i[/tex] is the change in gravitational potential energy
Solving for [tex]\Delta E_p[/tex],
[tex]\Delta E_p = W+W_f - \Delta K=80.4 J - 244 J - 1457.2 J = -1620.8 J[/tex]
(b) 2.97 m
The change in potential energy of the skateboarder can be written as
[tex]\Delta E_p = mg \Delta h[/tex]
where
m = 55.6 kg is the mass
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\Delta h[/tex] is the change in vertical height of the skateboarder
Solving for [tex]\Delta h[/tex],
[tex]\Delta h = \frac{\Delta E_p}{mg}=\frac{-1620.8 J}{(55.6 kg)(9.8 m/s^2)}=-2.97 m[/tex]
Where the negative sign means the skateboarder has moved downwards. Since we are interested only in the absolute value, the answer is
h = 2.97 m
a) The change in gravitational potential energy is 1536.92 joules.
b) The vertical height of the skateboarder changed in 2.819 meters.
How to apply the principle of energy conservation and work-energy theorem on a skateboarder
a) Based on the principle of energy conservation and work-energy theorem, the mechanical energy ([tex]E[/tex]), in joules, of the skateboarder is a combination of gravitational potential energy ([tex]U[/tex]), in joules, translational kinetic energy ([tex]K[/tex]), in joules, and work gain due to impact ([tex]W_{push}[/tex]) and work loss due to friction ([tex]W_{friction}[/tex]), in joules. The change in the gravitational potential energy is described below:
[tex]U_{1}+K_{1} + W_{push} = U_{2}+K_{2}+W_{friction}[/tex] (1)
[tex]U_{1}-U_{2} = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})+W_{friction}-W_{push}[/tex]
Where:
- [tex]m[/tex] - Mass of the skateboarder, in kilograms
- [tex]v[/tex] - Speed of the skateboarder, in meters per second
If we know that [tex]m = 55.6\,kg[/tex], [tex]v_{1} = 2.44\,\frac{m}{s}[/tex], [tex]v_{2} = 7.24\,\frac{m}{s}[/tex], [tex]W_{friction} = 244\,J[/tex] and [tex]W_{push} = 80.4\,J[/tex], then the change in gravitational potential energy is:
[tex]U_{1}-U_{2} = \frac{1}{2}\cdot (55.6\,kg)\cdot [\left(7.44\,\frac{m}{s} \right)^{2}-\left(2.44\,\frac{m}{s}} \right)^{2}] +244\,J - 80.4\,J[/tex]
[tex]U_{1}-U_{2} = 1536.92\,J[/tex]
The change in gravitational potential energy is 1536.92 joules. [tex]\blacksquare[/tex]
b) The change in height ([tex]h[/tex]), in meters, experimented by the skateboard is determined by definition of gravitational potential energy:
[tex]h = \frac{U_{1}-U_{2}}{m\cdot g}[/tex] (2)
Where [tex]g[/tex] is the gravitational acceleration, in meters per square second.
If we know that [tex]U_{1}-U_{2} = 1536.92\,J[/tex], [tex]m = 55.6\,kg[/tex] and [tex]g = 9.807 \,\frac{m}{s^{2}}[/tex], then the change in vertical height is:
[tex]h = \frac{1536.92\,J}{(55.6\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]h = 2.819\,m[/tex]
The vertical height of the skateboarder changed in 2.819 meters. [tex]\blacksquare[/tex]
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