A constant net force of 20 N is applied to a 32 kg car, causing it to speed up from 4.0 to 9.0 m/s. How long is the force applied?
This is chemistry???
If there is no friction then
a = F/m = 20/32 = 0.625 meters/ second^2
v = initial velocity + a t
9 = 4 + 0.625 t
t = 5 / 0.625 = 8 seconds
To find how long the force is applied, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of the object's mass and its acceleration:
Net force = mass x acceleration
In this case, the net force is 20 N, and the mass of the car is 32 kg. The car's initial speed is 4.0 m/s, and its final speed is 9.0 m/s. We need to find the acceleration first.
Using the equation for acceleration:
acceleration = (final speed - initial speed) / time
We know the initial and final speeds, but we don't know the time or acceleration. However, we can rearrange the equation to solve for time:
time = (final speed - initial speed) / acceleration
Now, let's find the acceleration:
acceleration = (final speed - initial speed) / time
acceleration = (9.0 m/s - 4.0 m/s) / time
acceleration = 5.0 m/s / time
Now we can substitute the value of acceleration into Newton's second law of motion:
Net force = mass x acceleration
20 N = 32 kg x 5.0 m/s / time
Now we can solve for time:
time = (32 kg x 5.0 m/s) / 20 N
time = 160 kg * m/s / 20 N
time = 8 s
Therefore, the force is applied for 8 seconds.