A(n) 23000 kg freight car is rolling along a
track at 2.8 m/s.
Calculate the time needed for a force of
1960 N to stop the car.
Answer in units of s.
Ft = mv
To calculate the time needed to stop the freight car, we can use the concept of Newton's second law of motion.
We know that the force acting on the car is 1960 N, and we need to find the time taken to bring the car to rest.
Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration:
F_net = m * a
In our case, the net force is equal to the force applied to stop the car, and the acceleration is the negative acceleration required to bring the car to rest.
The negative acceleration can be calculated using the equation:
a = -v^2 / (2 * d)
Where:
a = acceleration
v = initial velocity
d = distance traveled (which is unknown in this case)
Let's calculate the negative acceleration first:
a = -v^2 / (2 * d)
= - (2.8 m/s)^2 / (2 * d)
= - 7.84 m^2/s^2 / (2 * d)
= - 3.92 m^2/s^2 / d
Now, we can substitute this acceleration value into Newton's second law equation:
1960 N = (23000 kg) * (-3.92 m^2/s^2 / d)
Solving for d:
d = (23000 kg) * (-3.92 m^2/s^2) / 1960 N
= -46.4 m
We obtain a negative value for d, indicating that the car needs to travel a distance of 46.4 meters to come to a stop.
Now, we can use the kinematic equation to calculate the time taken to travel this distance at an initial velocity of 2.8 m/s:
d = v_i * t + (1/2) * a * t^2
Rearranging the equation and substituting the values we know:
t^2 - (v_i * t) / a - (2 * d) / a = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where:
a = 1/2 * a
b = -v_i
c = -2 * d
Substituting in the values:
t = (-(2.8 m/s) ± √((2.8 m/s)^2 - 4 * (1/2 * a) * (-2 * 46.4 m))) / (2 * 1/2 * a)
Simplifying:
t = (-2.8 m/s ± √(7.84 m^2/s^2 + 3.92 m/s^2 * 2 * 46.4 m)) / (2 * 3.92 m/s^2 / d)
Simplifying further:
t = (-2.8 m/s ± √(7.84 m^2/s^2 + 7.84 m^2/s^2 * 92.8 m)) / (7.84 m^2/s^2)
Calculating the square root and performing the addition and subtraction:
t = (-2.8 m/s ± √(7.84 m^2/s^2 + 728.192 m^3/s^2)) / (7.84 m^2/s^2)
t = (-2.8 m/s ± √(736.032 m^2/s^2)) / (7.84 m^2/s^2)
t = (-2.8 m/s ± √(736.032) m/s) / (7.84 m^2/s^2)
After calculating the square root, we obtain:
t = (-2.8 m/s ± 27.110 m/s) / (7.84 m^2/s^2)
Now, we consider both the positive and negative solutions:
t1 = (-2.8 m/s + 27.110 m/s) / (7.84 m^2/s^2)
= 24.310 m/s / (7.84 m^2/s^2)
= 3.104 s
t2 = (-2.8 m/s - 27.110 m/s) / (7.84 m^2/s^2)
= -29.910 m/s / (7.84 m^2/s^2)
= -3.818 s
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the time needed for a force of 1960 N to stop the car is approximately 3.104 seconds.
To calculate the time needed for a force to stop the car, we need to use the equation of motion that relates force, mass, and acceleration.
The equation of motion is given by Newton's second law:
F = ma
where F is the force applied, m is the mass of the car, and a is the acceleration.
To find the acceleration (a), we can use the following formula:
a = (final velocity - initial velocity) / time
In this case, the car is coming to a stop, so the final velocity (Vf) will be 0 m/s. The initial velocity (Vi) given in the problem is 2.8 m/s. We can now rearrange the formula to solve for time:
time = (final velocity - initial velocity) / acceleration
The force applied, F, is given in the problem as 1960 N, and the mass of the car, m, is given as 23,000 kg.
Let's substitute the values into the equation:
a = (0 - 2.8) / time
1960 = 23,000 * (0 - 2.8) / time
To find the time, we can now rearrange the equation:
time = 23,000 * (0 - 2.8) / 1960
Now we can calculate the time:
time = 23,000 * (-2.8/1960)
Simplifying the expression:
time ≈ -32.8571
Since time cannot be negative, we discard the negative sign:
time ≈ 32.8571 s
Therefore, the time needed for a force of 1960 N to stop the car is approximately 32.8571 seconds.