It takes a force of 53kN on the lead car of a 16-car passenger train with mass 9.1x 10^5 kg to pull it atconstant 45 m/s on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in pan (c) Would be needed to give the train an acceleration of 1.5m/s^2, at the instant the train has a speed of 45m/s on levels tracks? (d)How much more power to the lead car than calculated in pan (e) would be needed to move the train opposite of a 1.5% grade (slope anglea= arctan 0.015) at a constant 45m/s?
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Let's break down the given information step-by-step and solve the problem.
Step 1: Given information
Force (F) on the lead car = 53 kN
Mass (m) of the train = 9.1 x 10^5 kg
Velocity (v) of the train = 45 m/s
Acceleration (a) = 1.5 m/s^2
Grade (slope angle) = 1.5% (arctan 0.015)
Step 2: Calculate power required to pull the train at a constant speed of 45 m/s on level tracks.
The power required can be calculated using the formula: power (P) = force (F) x velocity (v).
(a) P = F x v
= (53 kN) x (45 m/s) [Note: Convert kN to N by multiplying by 1000]
= (53,000 N) x (45 m/s)
= 2,385,000 W
Therefore, the power required to pull the train at a constant speed of 45 m/s on level tracks is 2,385,000 W.
(b) Calculate the additional power required to overcome the acceleration of 1.5 m/s^2.
To calculate the additional power, we need to consider the force due to acceleration. The additional force required (F_extra) can be calculated using the formula: F_extra = m x a
F_extra = (9.1 x 10^5 kg) x (1.5 m/s^2)
= 1,365,000 N
To calculate the additional power (P_extra), we can use the equation: P_extra = F_extra x v
P_extra = (1,365,000 N) x (45 m/s)
= 61,425,000 W
Therefore, the additional power required to give the train an acceleration of 1.5 m/s^2 at the instant the train has a speed of 45 m/s on level tracks is 61,425,000 W.
(c) Calculate the power required to move the train up a 1.5% grade at a constant speed of 45 m/s.
To calculate the power required on a grade, we need to consider the force due to the grade. The force due to the grade (F_grade) can be calculated using the formula: F_grade = m x g x sin(theta)
F_grade = (9.1 x 10^5 kg) x (9.8 m/s^2) x sin(arctan(0.015))
To calculate the sin(arctan(0.015)), you can use a scientific calculator or an online calculator. The value is approximately 0.015.
F_grade = (9.1 x 10^5 kg) x (9.8 m/s^2) x 0.015
= 132,528 N
To calculate the power required (P_grade), we can use the equation: P_grade = F_grade x v
P_grade = (132,528 N) x (45 m/s)
= 5,966,760 W
Therefore, the power required to move the train up a 1.5% grade at a constant speed of 45 m/s is 5,966,760 W.
(d) Calculate the additional power required to move the train up the 1.5% grade.
To calculate the additional power, we need to subtract the power required on level tracks from the power required on the 1.5% grade.
Additional Power = Power on Grade - Power on Level Track
= 5,966,760 W - 2,385,000 W
= 3,581,760 W
Therefore, the additional power required to move the train up a 1.5% grade at a constant speed of 45 m/s is 3,581,760 W.
To answer the questions, we need to use the formulas related to power, force, and acceleration.
(a) The power required by the locomotive to the lead car can be calculated using the formula:
Power = Force x Velocity
We are given the force (53 kN) and velocity (45 m/s). However, we need to convert the force to Newtons before we can use it in the formula. 1 kN is equivalent to 1000 N, so 53kN is 53,000 N.
Substituting the values into the formula:
Power = 53,000 N x 45 m/s = 2,385,000 W
Therefore, the power required by the locomotive to the lead car is 2,385,000 Watts.
(b) To find how much more power is needed compared to the calculated power in part (a), we need to subtract the power calculated in part (a) from the new power. However, it seems there might be a typo in the question as it mentions "calculated in pan." Assuming it meant "part (a)" instead:
New Power = Force x Velocity
New Power = (53,000 N - 0 N) x 45 m/s = 2,385,000 W - 0 W = 2,385,000 W
The power required to the lead car remains the same, so there is no additional power needed.
(c) To calculate the power required to give the train an acceleration of 1.5 m/s^2, we need to use the formula:
Power = Force x Velocity
The force required can be determined using Newton's second law:
Force = Mass x Acceleration
We are given the mass of the train (9.1 x 10^5 kg) and the acceleration (1.5 m/s^2).
Substituting the values:
Force = 9.1 x 10^5 kg x 1.5 m/s^2 = 1.365 x 10^6 N
Now we can calculate the power:
Power = 1.365 x 10^6 N x 45 m/s = 61,425,000 W
Therefore, the power required to give the train an acceleration of 1.5 m/s^2 is 61,425,000 Watts.
(d) Similar to part (b), it seems there might be a typo in the question as it mentions "calculated in pan." Assuming it meant "part (c)" instead:
New Power = Force x Velocity
New Power = (1.365 x 10^6 N - 53,000 N) x 45 m/s = 61,425,000 W - 2,385,000 W = 59,040,000 W
The additional power required compared to part (c) is 59,040,000 - 61,425,000 = -2,385,000 W (negative value indicates less power is needed).
(e) Moving the train on a 1.5% grade (slope angle a = arctan 0.015) at a constant velocity of 45 m/s requires overcoming the force of gravity acting on the train. The force required to overcome the gravitational force can be calculated using:
Force = Mass x Gravitational Acceleration x sin(a)
where, a is the slope angle.
We are given the mass of the train (9.1 x 10^5 kg), the gravitational acceleration (9.8 m/s^2), and the slope angle (arctan(0.015)).
Substituting the values:
Force = 9.1 x 10^5 kg x 9.8 m/s^2 x sin(arctan(0.015))
Note: When dealing with trigonometric functions, it's better to convert the slope angle to radians.
Force = 9.1 x 10^5 kg x 9.8 m/s^2 x sin(0.015 radians)
Now, we can calculate the power:
Power = Force x Velocity
Power = (9.1 x 10^5 kg x 9.8 m/s^2 x sin(0.015)) x 45 m/s
Therefore, the power required to move the train opposite a 1.5% grade at a constant velocity of 45 m/s can be calculated using the above formula.
1) 0.5mv1^2 - W = 0.5mv2^2
v1 = initial speed
v2 = final speed = 0
re-arrange
0.5mv1^2 = W
Work in this case is defined as umd
u = coeffecient of friction
m = mass
d = distance it travelled
since coeffecient of friction is a function of distance, work is changing, so you need to integrate
0.5mv^2 = m* integral from 0 to x of (u*d) + 0.6*m*(d-12.5)
(do a calculation to see if it stops before or after to include the second term for work.
mass is useless, and can be eliminated. final format looks like
0.5v1^2 = integral from 0 to x of (u*d) + 0.6*(d-12.5)
find u in terms of distance ((0.6-0.1)/12.5)*x + 0.1
b) if it stops before 12.5, you solved for distance previously, plug it into u eqn to get u.
c) .5v^2