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?
frick
a=gsin(theta)
a=9.8 sin 25°
a= 0.43
Uf=Ui +at
Uf= 0 + a(10)
Uf=0 +4.3
Uf=4.3 m/s
To calculate the speed of the boxcar after 10 seconds, we can use the principles of physics, specifically Newton's second law of motion.
1. First, let's calculate the force acting on the boxcar due to the grade. We know that the mass of the boxcar is 200 tons, which is equal to 200,000 kilograms (since 1 ton = 1000 kg). The force due to gravity can be calculated using the equation:
F = m * g * sin(theta)
Where F is the force, m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the grade (2.5 degrees).
F = 200,000 kg * 9.8 m/s^2 * sin(2.5 degrees)
2. Calculate the force acting on the boxcar:
F = 200,000 kg * 9.8 m/s^2 * sin(2.5) = 8,416 N (rounded to three decimal places)
3. With the force acting on the boxcar, we can now calculate its acceleration using Newton's second law of motion:
F = m * a
Rearranging the equation to solve for acceleration:
a = F / m
a = 8,416 N / 200,000 kg = 0.04208 m/s^2 (rounded to five decimal places)
4. Now that we have the acceleration, we can use the kinematic equation to find the final velocity:
v = u + a * t
Where v is the final velocity, u is the initial velocity (which is 0 m/s since the boxcar starts at rest), a is the acceleration, and t is the time (10 seconds).
v = 0 m/s + 0.04208 m/s^2 * 10 s = 0.4208 m/s (rounded to four decimal places)
So, the boxcar will have a speed of approximately 0.4208 m/s after 10 seconds.
To find the speed of the boxcar after 10 seconds, we can use the principles of physics, specifically motion and force equations.
First, we need to determine the forces acting on the boxcar. In this case, we have three forces to consider: the force due to gravity, the normal force, and the force of friction.
1. Force due to gravity:
The force due to gravity can be calculated using the formula:
F_gravity = m * g
where m is the mass of the boxcar and g is the acceleration due to gravity. In this case, the mass of the boxcar is given as 200 tons. Since 1 ton is equal to 1000 kg, the mass in kilograms would be:
m = 200 tons * 1000 kg/ton = 200000 kg
The acceleration due to gravity can be assumed to be 9.8 m/s^2.
Evaluating the formula, we get:
F_gravity = 200000 kg * 9.8 m/s^2 = 1.96 * 10^6 N
2. Normal force:
The normal force is the force exerted by a surface perpendicular to the surface. In this case, the boxcar is on a 2.5 degree grade. Therefore, the normal force can be calculated using:
F_normal = m * g * cos(theta)
where theta is the angle of the grade, which is 2.5 degrees in this case. We need to convert the angle to radians:
theta_rad = (2.5 degrees * pi) / 180 = 0.0436 rad
Evaluating the formula, we get:
F_normal = 200000 kg * 9.8 m/s^2 * cos(0.0436 rad) = 1.95974 * 10^6 N
3. Force of friction:
In this case, the track is considered to be frictionless. Therefore, there is no force of friction acting on the boxcar.
Since there is no net force acting horizontally, the boxcar will undergo constant acceleration down the grade.
We can use the following equation to find the speed after 10 seconds:
v = u + a * t
where v is the final velocity, u is the initial velocity (which is 0 m/s since the boxcar is at rest initially), a is the acceleration, and t is the time.
Since the boxcar undergoes constant acceleration, we can use the formula:
a = (F_gravity - F_normal) / m
Evaluating the formula, we get:
a = (1.96 * 10^6 N - 1.95974 * 10^6 N) / 200000 kg = 1.3 N/kg
Plugging in the values into the final velocity equation:
v = 0 m/s + 1.3 m/s^2 * 10 s = 13 m/s
Therefore, the boxcar will have a speed of 13 m/s after 10 seconds.