A boxcar of mass 250 tons at rest becomes uncoupled on a 4.0° grade. If the track is considered to be frictionless, what speed does the boxcar have after 11 seconds?
v = at
a = g sin4°
Why did the boxcar go to the gym? It wanted to "train" on a grade! Now, let's calculate its speed, shall we?
Given:
Mass of the boxcar (m): 250 tons = 250,000 kg
Grade of the track (θ): 4.0°
Time (t): 11 seconds
First, let's calculate the gravitational force acting on the boxcar:
Force = Mass × Acceleration due to Gravity
F_gravity = m × g
Since the track is frictionless, there are no other forces acting on the boxcar. Therefore, the gravitational force will be the net force.
So, m × g = Net Force
To calculate the Net Force in the direction of the grade:
Net Force = (m × g) × sin(θ)
Now, let's determine the acceleration of the boxcar along the grade:
Acceleration = Net Force / m
Using the equations of motion:
Final speed = Initial speed + (Acceleration × Time)
But since the boxcar is initially at rest, the initial speed is zero.
So, Final speed = (Acceleration × Time)
Plugging in the values:
Final speed = (Acceleration × 11)
Now, let's calculate!
Oh wait, before that, here's a quick joke for you: Why was the math book sad? Because it had too many problems! Okay, back to business now.
Since the gravitational force is acting in the direction of the grade, the equation becomes:
F_gravity = m × g × sin(θ)
Substituting the values:
F_gravity = 250,000 kg × 9.8 m/s² × sin(4.0°)
Calculating F_gravity, we find:
F_gravity ≈ 41,649 N
Now, let's calculate the acceleration:
Acceleration = F_gravity / m
Acceleration = 41,649 N / 250,000 kg
Calculating the acceleration, we find:
Acceleration ≈ 0.1666 m/s²
Finally, let's calculate the final speed:
Final speed = Acceleration × Time
Final speed = 0.1666 m/s² × 11 s
Calculating the final speed, we find:
Final speed ≈ 1.833 m/s
So, after 11 seconds, the boxcar will have a speed of approximately 1.833 m/s.
To find the speed of the boxcar after 11 seconds, we can use the equations of motion considering the net force acting on the boxcar.
First, let's determine the weight force acting on the boxcar. The weight force is given by:
Weight = mass * gravity
Where the mass of the boxcar is 250 tons, and the acceleration due to gravity is approximately 9.8 m/s².
Weight = 250 tons * 9.8 m/s² = 2450 kN
Next, we need to determine the component of the weight force acting along the inclined plane. This component is given by:
Force_parallel = Weight * sin(θ)
Where θ is the angle of the incline, which is 4.0° in this case.
Force_parallel = 2450 kN * sin(4.0°) = 169.45 kN
Since the track is considered frictionless, the only force acting on the boxcar along the incline is the parallel component of gravity. Therefore, this force is responsible for the acceleration of the boxcar.
The force acting on the boxcar can be related to its acceleration using Newton's second law:
Force = mass * acceleration
Rearranging the equation, we can find the acceleration:
acceleration = Force / mass
acceleration = 169.45 kN / (250 tons * 9.8 m/s²)
Let's convert the mass from tons to kilograms:
mass = 250 tons * (1000 kg/ton) = 250000 kg
acceleration = 169.45 kN / (250000 kg * 9.8 m/s²)
acceleration ≈ 0.069 m/s²
Now, we can find the speed of the boxcar after 11 seconds. We'll use the equation of motion:
final velocity = initial velocity + (acceleration * time)
Since the boxcar is initially at rest (initial velocity = 0), the equation simplifies to:
final velocity = acceleration * time
final velocity = 0.069 m/s² * 11 s
final velocity ≈ 0.759 m/s
Therefore, the boxcar has a speed of approximately 0.759 m/s after 11 seconds.
To find the speed of the boxcar after 11 seconds, we can use the equations of motion. The key equation we will use is:
v = u + at
Where:
v is the final velocity (speed)
u is the initial velocity (initial speed, in this case, 0 m/s, since the boxcar is at rest)
a is the acceleration
t is the time
First, let's find the acceleration of the boxcar on the 4.0° grade. The force acting on the boxcar down the incline is its weight (mg) where m is the mass of the boxcar and g is the acceleration due to gravity. On the incline, the force can be split into two components: the parallel component (mg*sin(θ)) and the perpendicular component (mg*cos(θ)).
The perpendicular component cancels out since it is balanced by the normal force of the incline. So, only the parallel component, mg*sin(θ), contributes to the acceleration. We can calculate the acceleration using the equation:
a = g * sin(θ)
First, let's convert the mass of the boxcar from tons to kilograms. Since 1 ton is equal to 1000 kilograms:
250 tons = 250 * 1000 kg = 250,000 kg
Now, let's calculate the acceleration:
a = 9.8 m/s^2 * sin(4.0°)
Using a calculator, we find that sin(4.0°) is approximately 0.069756.
a ≈ 9.8 m/s^2 * 0.069756 ≈ 0.683 m/s^2
Now, we can use the equation of motion to find the final velocity:
v = u + at
v = 0 m/s + (0.683 m/s^2) * (11 s)
v ≈ 7.513 m/s
Therefore, the boxcar will have a speed of approximately 7.513 m/s after 11 seconds.