11) ![1.04\cdot 10^7 J](https://tex.z-dn.net/?f=1.04%5Ccdot%2010%5E7%20J)
12) ![1.04\cdot 10^7 J](https://tex.z-dn.net/?f=1.04%5Ccdot%2010%5E7%20J)
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
![PE=mgh](https://tex.z-dn.net/?f=PE%3Dmgh)
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)
(acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
![PE=(8325)(9.8)(127)=1.04\cdot 10^7 J](https://tex.z-dn.net/?f=PE%3D%288325%29%289.8%29%28127%29%3D1.04%5Ccdot%2010%5E7%20J)
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:
![KE_t + PE_t = KE_b + PE_b](https://tex.z-dn.net/?f=KE_t%20%2B%20PE_t%20%3D%20KE_b%20%2B%20PE_b)
where
is the kinetic energy at the top
is the potential energy at the top
is the kinetic energy at the bottom
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
![KE_t=0](https://tex.z-dn.net/?f=KE_t%3D0)
Also, at the bottom the height is zero, so the potential energy is zero
![PE_b=0](https://tex.z-dn.net/?f=PE_b%3D0)
Therefore, we find:
![KE_b=PE_t=1.04\cdot 10^7 J](https://tex.z-dn.net/?f=KE_b%3DPE_t%3D1.04%5Ccdot%2010%5E7%20J)
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
![KE=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
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
![KE=1.04\cdot 10^7 J](https://tex.z-dn.net/?f=KE%3D1.04%5Ccdot%2010%5E7%20J)
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
![v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2KE%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%281.04%5Ccdot%2010%5E7%29%7D%7B8325%7D%7D%3D50.0%20m%2Fs)
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
![KE_1 + PE_1 = KE_2 + PE_2](https://tex.z-dn.net/?f=KE_1%20%2B%20PE_1%20%3D%20KE_2%20%2B%20PE_2)
where
is the kinetic energy at the top of the 1st hill
is the potential energy at the top of the 1st hill
is the kinetic energy at the top of the 2nd hill
is the potential energy at the top of the 2nd hill
From the previous questions, we know that
![KE_1=0](https://tex.z-dn.net/?f=KE_1%3D0)
and
![PE_1=1.04\cdot 10^7 J](https://tex.z-dn.net/?f=PE_1%3D1.04%5Ccdot%2010%5E7%20J)
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
![PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J](https://tex.z-dn.net/?f=PE_2%3Dmgh%3D%288325%29%289.8%29%2839%29%3D3.2%5Ccdot%2010%5E6%20J)
So, the kinetic energy at the second hill is
![KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J](https://tex.z-dn.net/?f=KE_2%3DPE_1-PE_2%3D1.04%5Ccdot%2010%5E7%20-%203.2%5Ccdot%2010%5E6%20%3D7.2%5Ccdot%2010%5E6%20J)
And so, the speed is
![v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2KE_2%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%287.2%5Ccdot%2010%5E6%29%7D%7B8325%7D%7D%3D41.6%20m%2Fs)