11) [tex]1.04\cdot 10^7 J[/tex]
12) [tex]1.04\cdot 10^7 J[/tex]
13) 50.0 m/s
14) 41.6 m/s
Explanation:
11)
The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by
[tex]PE=mgh[/tex]
where
m is the mass of the object
g is the acceleration due to gravity
h is the height relative to the ground
Here in this problem, when the train is at the top, we have:
m = 8325 kg (mass of the train + riders)
[tex]g=9.8 m/s^2[/tex] (acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
[tex]PE=(8325)(9.8)(127)=1.04\cdot 10^7 J[/tex]
12)
According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:
[tex]KE_t + PE_t = KE_b + PE_b[/tex]
where
[tex]KE_t[/tex] is the kinetic energy at the top
[tex]PE_t[/tex] is the potential energy at the top
[tex]KE_b[/tex] is the kinetic energy at the bottom
[tex]PE_b[/tex] is the potential energy at the bottom
The kinetic energy is the energy due to motion; since the train is at rest at the top, we have
[tex]KE_t=0[/tex]
Also, at the bottom the height is zero, so the potential energy is zero
[tex]PE_b=0[/tex]
Therefore, we find:
[tex]KE_b=PE_t=1.04\cdot 10^7 J[/tex]
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is the speed of the object
From question 12), we know that the kinetic energy of the train at the bottom is
[tex]KE=1.04\cdot 10^7 J[/tex]
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
[tex]v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s[/tex]
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
[tex]KE_1 + PE_1 = KE_2 + PE_2[/tex]
where
[tex]KE_1[/tex] is the kinetic energy at the top of the 1st hill
[tex]PE_1[/tex] is the potential energy at the top of the 1st hill
[tex]KE_2[/tex] is the kinetic energy at the top of the 2nd hill
[tex]PE_2[/tex] is the potential energy at the top of the 2nd hill
From the previous questions, we know that
[tex]KE_1=0[/tex]
and
[tex]PE_1=1.04\cdot 10^7 J[/tex]
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
[tex]PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J[/tex]
So, the kinetic energy at the second hill is
[tex]KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J[/tex]
And so, the speed is
[tex]v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s[/tex]
