1) 9.4 m/s
First of all, we can calculate the work done by the horizontal force, given by
W = Fd
where
F = 34.6 N is the magnitude of the force
d = 12.9 m is the displacement of the cart
Solving ,
W = (34.6 N)(12.9 m) = 446.3 J
According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:
[tex]W=K_f - K_i[/tex]
Since the cart was initially at rest, [tex]K_i = 0[/tex], so
[tex]W=K_f = \frac{1}{2}mv^2[/tex] (1)
where
m is the of the cart
v is the final speed
The mass of the cart can be found starting from its weight, [tex]F_g = 99.5 N[/tex]:
[tex]m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg[/tex]
So solving eq.(1) for v, we find the final speed of the cart:
[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s[/tex]
2) [tex]2.51\cdot 10^7 J[/tex]
The work done on the train is given by
W = Fd
where
F is the magnitude of the force
d is the displacement of the train
In this problem,
[tex]F=4.28 \cdot 10^5 N[/tex]
[tex]d=586 m[/tex]
So the work done is
[tex]W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J[/tex]
3) [tex]2.51\cdot 10^7 J[/tex]
According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:
[tex]W=\Delta K = K_f - K_i[/tex]
where
W is the work done
[tex]\Delta K[/tex] is the change in kinetic energy
Therefore, the change in kinetic energy is
[tex]\Delta K = W = 2.51\cdot 10^7 J[/tex]
4) 37.2 m/s
According to the work-energy theorem,
[tex]W=\Delta K = K_f - K_i[/tex]
where
[tex]K_f[/tex] is the final kinetic energy of the train
[tex]K_i = 0[/tex] is the initial kinetic energy of the train, which is zero since the train started from rest
Re-writing the equation,
[tex]W=K_f = \frac{1}{2}mv^2[/tex]
where
m = 36300 kg is the mass of the train
v is the final speed of the train
Solving for v, we find
[tex]v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s[/tex]
