A 2010 kg car traveling to the west at 18.5
m/s slows down uniformly under a force of
8880 N to the east.
a) How much force would be required to
cause the same acceleration on a car of mass
3120 kg?
How far would the car move before stop-
ping?
Answer in units of m
To find the force required to cause the same acceleration on a car of mass 3120 kg, we can use Newton's second law of motion:
Force = mass * acceleration
Given:
Mass of the new car (m1) = 3120 kg
Mass of the initial car (m2) = 2010 kg
Initial velocity (v2) = 18.5 m/s
Final velocity (v1) = 0 m/s
Acceleration can be calculated using the equation:
a = (v1 - v2) / t
Since the car is coming to a stop, the time (t) taken will be the same for both cars.
Let's calculate the acceleration first:
a = (0 - 18.5) / t
Now, we can equate the forces for both cars:
Force1 = mass1 * acceleration
Force2 = mass2 * acceleration
Since Force2 is given as 8880 N:
Force1 = 3120 kg * acceleration
8880 N = 2010 kg * acceleration
Dividing both sides of the equation by the mass of the new car:
acceleration = 8880 N / 2010 kg
Substituting this value of acceleration back into the equation for the force required for the new car:
Force1 = 3120 kg * (8880 N / 2010 kg)
Now, we can find the force required for the new car:
Force1 = 13680 N
Therefore, the force required to cause the same acceleration on a car of mass 3120 kg would be 13680 N.
To find how far the car would move before stopping, we can use the equation:
v^2 = v0^2 + 2a * x
Since the final velocity is 0, the equation simplifies to:
0 = (v0)^2 + 2a * x
Rearranging the equation to solve for x (distance):
x = (-v0)^2 / (2a)
Now we can substitute the values into the equation:
x = (-18.5 m/s)^2 / (2 * acceleration)
Substituting the value of acceleration:
x = (-18.5 m/s)^2 / (2 * (8880 N / 2010 kg))
x = 31.25 m
Therefore, the car would move approximately 31.25 meters before coming to a stop.
To find the force required to cause the same acceleration on a car of mass 3120 kg, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given:
Mass of the first car (m1) = 2010 kg
Velocity of the first car (v1) = 18.5 m/s (west)
Force acting on the first car (F1) = 8880 N (east)
Acceleration is the rate at which velocity changes over time. Since the first car slows down uniformly, its acceleration (a1) can be calculated using the formula:
a1 = (v1 - 0) / t
Where t is the time taken for the car to stop. However, we do not have the value of t.
To find the time (t), we can use the equation of motion:
v1 = u + at
Where u is the initial velocity (18.5 m/s) and v1 is the final velocity (0 m/s).
0 = 18.5 + a1 * t
a1 * t = -18.5
t = -18.5 / a1
Using the equation F = m * a, we can calculate the acceleration (a1) of the first car:
F1 = m1 * a1
a1 = F1 / m1
a1 = 8880 N / 2010 kg
Now, we can calculate the time (t) taken for the car to stop:
t = -18.5 / (8880 N / 2010 kg)
Once we have calculated the time taken to stop, we can find the acceleration (a2) for the second car using the formula:
a2 = (v2 - 0) / t
Given:
Mass of the second car (m2) = 3120 kg
Velocity of the second car (v2) = 0 m/s (since it is stopped)
Now, we can calculate the force (F2) required to cause the same acceleration on the second car:
F2 = m2 * a2
Finally, to find the distance traveled before stopping, we can use the equation:
d = (v1^2) / (2 * a1)
Please provide the value of the final velocity of the first car (v1) in order to continue with the calculations.