Answer:
A) E = 2550 J
B) K = 1325 J
C) t = 5,05 s
Explanation:
A) The total mechanical energy is given by the sum of the gravitational potential energy and the kinetic energy of the body:
[tex]E=U+K=mgh+\frac{1}{2}mv^2[/tex] (1)
m: mass of the body = 2,0 kg
g: gravitational acceleration = 9,8 m/s^2
h: height = 125 m
v: initial velocity of the body = 10 m/s
You replace the values of all variables h, m, g and v in the equation (1):
[tex]E=(2,0kg)(9,8m/s^2)(125m)+\frac{1}{2}(2,0kg)(10m/s)^2=2550\ J[/tex]
the total mechanical energy is 2550 J
B) The kinetic energy of the corp, when it is at a height of h/2 is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
[tex]v=\sqrt{(v_x)^2+(v_y)^2}[/tex]
The x component of the velocity is constant in the complete trajectory, which is the initial velocity, that is, vo = vx
The y component is given by:
[tex]v_y^2=v_{oy}^2+2gy[/tex]
voy: vertical initial velocity = 0m/s
y: height = h/2 = 125/2 = 62.5 m
[tex]v_y=\sqrt{2g\frac{h}{2}}=\sqrt{2(9.8m/s^2)(62.5m)}=35m/s[/tex]
Then, you can calculate the velocity of the body and next, you can calculate the kinetic energy:
[tex]v=\sqrt{(10m/s)^2+(35m/s)^2}=36,40\frac{m}{s}\\\\K=\frac{1}{2}(2,0kg)(36,40m/s)^2=1325\ J[/tex]
C) The time that body takes in all its trajectory is:
[tex]t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(125m)}{9,8m/s^2}}=5,05s[/tex]
