A boxcar of mass 200 tons at rest becomes uncoupled on a 2.5 degree grade. If the track is considered to be frictionless, what speed does the boxcar have after 10 seconds?

First compute the acceleration. It will be a = g * sin 2.5

Then compute how far it travels after 10 seconds. X = (1/2) a t^2

thank you so much. i never knew it could be so simple

To find the speed of the boxcar after 10 seconds, we can use the concept of mechanical energy conservation. The mechanical energy of a system is the sum of potential energy and kinetic energy.

1. Since the track is considered to be frictionless, there is no loss of mechanical energy due to friction.

2. The only energy change in this case is the conversion of gravitational potential energy (due to the boxcar's position on the grade) into kinetic energy (as the boxcar gains speed).

3. The change in potential energy is equal to the change in kinetic energy. Mathematically, this can be expressed as:

mgh = 0.5mv^2,

where m is the mass of the boxcar, g is the acceleration due to gravity, h is the change in elevation, and v is the velocity.

4. On a 2.5-degree grade, the change in elevation can be calculated using trigonometry:

h = length of the grade * sin(2.5 degrees).

Since the length of the grade is not given, we can assume it doesn't affect the final velocity and use a hypotenuse of 1.

h = 1 * sin(2.5 degrees).

5. We can convert the mass of the boxcar from tons to kilograms:

1 ton = 1000 kg,
Therefore, the mass of the boxcar is 200 tons * 1000 kg/ton = 200,000 kg.

6. The acceleration due to gravity, g, is approximately 9.8 m/s^2.

7. Substituting the given values into the equation, we can solve for v:

(200,000 kg)(9.8 m/s^2)(sin(2.5 degrees)) = 0.5(200,000 kg)v^2.

8. Simplifying the equation:

v^2 = (2 * (200,000 kg)(9.8 m/s^2)(sin(2.5 degrees)) ) / 200,000 kg.

v^2 = (2 * 9.8 m/s^2 * sin(2.5 degrees)).

9. Calculating the right side of the equation:

v^2 = (2 * 9.8 m/s^2 * sin(2.5 degrees)).
v^2 = 0.85682 m^2/s^2.

10. Taking the square root to find v, the velocity of the boxcar after 10 seconds:

v ≈ √0.85682 m^2/s^2 ≈ 0.9268 m/s.

Therefore, the boxcar has a speed of approximately 0.9268 m/s after 10 seconds.

To find the speed of the boxcar after 10 seconds, we can use the principles of motion and energy.

Step 1: Determine the gravitational force acting on the boxcar.
The gravitational force is given by the formula F = m * g, where m is the mass of the boxcar and g is the acceleration due to gravity (approximately 9.8 m/s²). Converting the mass of the boxcar from tons to kilograms, we have:
m = 200 tons * (1000 kg/1 ton) = 200,000 kg

F = 200,000 kg * 9.8 m/s² = 1,960,000 N

Step 2: Calculate the component of the gravitational force on the incline.
The component of the force acting parallel to the incline is given by the formula F_parallel = F * sin(θ), where θ is the angle of the incline.
θ = 2.5 degrees, however, we need to convert it to radians.
θ_radians = 2.5 degrees * (π/180 degrees) = 0.0436 radians

F_parallel = 1,960,000 N * sin(0.0436 radians) = 1,960,000 N * 0.0436 = 85,616 N

Step 3: Calculate the net force on the boxcar.
Since the track is considered to be frictionless, there is no frictional force acting on the boxcar. Therefore, the net force is equal to the component of the gravitational force parallel to the incline:
Net Force = F_parallel = 85,616 N

Step 4: Calculate the acceleration of the boxcar.
The net force is equal to the mass of the object multiplied by its acceleration (F = m * a). Rearranging the formula, we get:
a = F / m

a = 85,616 N / 200,000 kg = 0.42808 m/s²

Step 5: Calculate the final speed of the boxcar after 10 seconds.
We can use the formula for linear motion:
v = u + a * t

where:
v = final velocity (speed)
u = initial velocity (since the boxcar starts from rest, u = 0)
a = acceleration (0.42808 m/s²)
t = time (10 seconds)

v = 0 + 0.42808 m/s² * 10 s = 4.2808 m/s or approximately 4.3 m/s

Therefore, the speed of the boxcar after 10 seconds is approximately 4.3 m/s.