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?
To find the speed of the boxcar after 10 seconds, we need to consider the forces acting on it. In this case, the only force that affects the boxcar's motion is the force due to gravity pulling it down the slope.
To start, we need to calculate the gravitational force acting on the boxcar. The formula for gravitational force is given by:
F_gravity = m * g
Where:
- F_gravity is the gravitational force (in Newtons)
- m is the mass of the boxcar (in kilograms)
- g is the acceleration due to gravity (approximately 9.81 m/s^2)
First, we need to convert the mass of the boxcar from tons to kilograms. Since 1 ton is equal to 1000 kilograms, the mass of the boxcar is:
m = 200 tons * 1000 kg/ton = 200,000 kg
Substituting the values into the formula, we get:
F_gravity = 200,000 kg * 9.81 m/s^2 ≈ 1,962,000 N
Next, we need to calculate the force component acting parallel to the slope. This force can be calculated by multiplying the gravitational force by the sine of the angle of the slope:
F_parallel = F_gravity * sin(θ)
Where:
- F_parallel is the force parallel to the slope
- θ is the angle of the slope (2.5 degrees)
Converting the angle from degrees to radians (since the sine function expects radians), we have:
θ_radians = θ * π/180 = 2.5 degrees * π/180 ≈ 0.0436 radians
Substituting the values into the formula, we get:
F_parallel = 1,962,000 N * sin(0.0436) ≈ 1,498 N
Now that we have the force parallel to the slope, we can calculate the acceleration of the boxcar using Newton's second law, which states:
F = m * a
Where:
- F is the force acting on the object (in Newtons)
- m is the mass of the object (in kilograms)
- a is the acceleration of the object (in meters per second squared)
Solving for acceleration, we have:
a = F_parallel / m
Substituting the values into the formula, we get:
a = 1,498 N / 200,000 kg ≈ 0.00749 m/s^2
Finally, we can calculate the final velocity of the boxcar after 10 seconds using the equation of motion:
v = u + a * t
Where:
- v is the final velocity of the object (in meters per second)
- u is the initial velocity of the object (in this case, 0 m/s since the boxcar is at rest)
- a is the acceleration of the object (in meters per second squared)
- t is the time interval (in seconds)
Substituting the values into the formula, we get:
v = 0 + 0.00749 m/s^2 * 10 s ≈ 0.0749 m/s
Therefore, the speed of the boxcar after 10 seconds on the 2.5 degree grade is approximately 0.0749 m/s.