Calculating Average Net Force of 810-kg Car

In summary, using the equations for displacement and acceleration, it was determined that the average net force acting on the car was 2460.375 N. This was found by first calculating the time and acceleration of the car and then using the formula for force. This answer was confirmed by also using the equations for kinetic energy and work.
  • #1
brncsfns5621
22
0
An 810-kg car accelerates from rest to 27m/s in a distance of 120m. What is the magnitude of the average net force acting on the car?

120m / 27m/s= 4.4 s

F=ma
=810kg * 4.4s
= 3600 N

This is not one of my choices. What did I do wrong?
 
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  • #2
brncsfns5621 said:
An 810-kg car accelerates from rest to 27m/s in a distance of 120m. What is the magnitude of the average net force acting on the car?

120m / 27m/s= 4.4 s

F=ma
=810kg * 4.4s
= 3600 N

This is not one of my choices. What did I do wrong?

You can not apply t = s / v, because the car is accelerating. The velocity of the car is given with 27 = a * t. The second equation is the displacement of the car, 120 = 1/2*a*t^2. Now, plug in the first equation into the second one, and get the time and acceleration. Then you can calculate the average force acting on the car.
 
  • #3
So, using your equations I get:

120m = 1/2 (27m/s)^2
120m = 1/2 (729m^2/s^2)
120m = 364m^2/s^2
rearrange--> s^2 = 364m^2/120m
s^2 = 3.03m

This isn't right...
 
  • #4
Anyone have any info on this?
 
  • #5
brncsfns5621 said:
Anyone have any info on this?

You have two equations:

(1)... 27 = a*t
(2)... 120 = 1/2*a*t^2

From (1) you obtain a = 27/t. Putting that into (2) gives t = 8.89 seconds, which implies (from (1) ) a = 3.04 m/s^2. Hence, F = m*a implies F = 810 * 3.04 = 2462.4 [N]. Do not mix units with variables. You do not have tu put units (such as m/s, for example) into your equations. In other words, you wrote
brncsfns5621 said:
...s^2 = 3.03m

This isn't right...
which reads: 'Seconds squared equals 3.03 meters.' Does not make any sense, right?
 
Last edited:
  • #6
WOW! I brain farted that one. For some reason I was thinking to replace the "a*t" with 27. Don't know why... Thank you for your help, I really appreciate it.
 
  • #7
solve the problem

a = ?
a = v/t
solve t using d=.5(v-V)t
t= 8.8sec

sub in the equation
a = (27m/s)/8.8sec
a = 3.06m/s^2

F= mass*acceleration
810kg*3.06m/s^2 = 2485N = 2500N
 
  • #8
KE = 1/2 * m v^2
KE = (1/2) (810) (27^2)
W= deltaKE
W = (1/2) (810) (27^2)
W = force * distance
(1/2)(810)(27^2) = 120*F
F= [(1/2)(810)(27^2)] / (120)
F= 2460.375 N
 

1. How do you calculate the average net force of a 810-kg car?

The average net force of a 810-kg car can be calculated by dividing the car's mass by its acceleration. This is represented by the formula F = m x a, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration would be the change in velocity over time, also known as the car's deceleration.

2. What is the unit of measurement for the average net force of a 810-kg car?

The unit of measurement for the average net force of a 810-kg car would be in Newtons (N). This is the standard unit for force in the International System of Units (SI).

3. How does the average net force affect the motion of a 810-kg car?

The average net force directly affects the motion of a 810-kg car. According to Newton's Second Law of Motion, the net force acting on an object is equal to its mass multiplied by its acceleration. Therefore, a larger net force will result in a greater acceleration and a smaller net force will result in a smaller acceleration. This means that the average net force can determine how quickly the car will accelerate or decelerate.

4. Can the average net force of a 810-kg car be negative?

Yes, the average net force of a 810-kg car can be negative. This would occur when the car is decelerating, or slowing down. In this case, the acceleration would have a negative value, which would result in a negative net force.

5. How can the average net force of a 810-kg car be applied in real-life situations?

The average net force of a 810-kg car can be applied in real-life situations to determine the braking distance needed for a car to come to a complete stop. By calculating the average net force, we can estimate the car's deceleration and use this information to determine the distance required to stop the car at a certain speed. This can be useful for drivers to maintain a safe following distance and to prevent accidents.

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