Answer:
C. 39 m/s
Explanation:
First we need to calculate the total force required to move the train along the inclined plane. So, it is clear that the work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present:
Force = F = mg Sin θ
where,
m = mass of train = 3.3 x 10⁶ kg
g = 9.8 m/s²
θ = Angle of Inclination = 0.64°
Therefore,
F = (3.3 x 10⁶ kg)(9.8 m/s²)Sin 0.64°
F = 3.612 x 10⁵ N
Now, the formula for power is:
P = FV
V = P/F
where,
V = Velocity of Train = ?
P = Power of Engine = 14 MW = 1.4 x 10⁷ W
Therefore,
V = 1.4 x 10⁷ W/3.612 x 10⁵ N
V = 38.75 m/s
which is approximately equal to:
C. 39 m/s
The speed of train is 39 m/s. Hence, option (C) is correct.
Given data:
The mass of train is, [tex]m = 3.3 \times 10^{6} \;\rm kg[/tex].
The angle of inclination is, [tex]\theta = 0.64^ {\circ}[/tex].
The useful power output value is, [tex]P= 14 \;\rm MW = 14 \times 10^{6} \;\rm W[/tex].
The work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present. So, the total force required to move the train along the inclined plane is given as,
[tex]F = mg sin \theta\\\\F = 3.3 \times 10^{6} \times 9.8 \times sin0.64\\\\F = 361233.75 \;\rm N[/tex]
Now, use the formula of the power to obtain the value of speed as,
[tex]P = F \times v\\\\14 \times 10^{6} =361233.75 \times v\\\\v \approx 38.75 \;\rm m/s[/tex]
Thus, the speed of train is 39 m/s. Hence, option (C) is correct.
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