a car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0s. What is the net force applied to the car?
Assuming constant acceleration,
s = 1/2 at^2, so
1/2 * a * 3^2 = 41
a = 9.111
And, we all know that F = ma
To find the net force applied to the car, we can use Newton's second law of motion:
Force (F) = mass (m) * acceleration (a)
First, let's calculate the acceleration of the car using the formula:
Acceleration (a) = change in velocity (Δv) / time (t)
Given that the car starts from rest, the change in velocity (Δv) would be equal to the final velocity (v) since the initial velocity is 0.
Using the formula:
Velocity (v) = distance (d) / time (t)
We can determine the final velocity:
v = 41 m / 3.0 s
v ≈ 13.67 m/s
Now, we can calculate the acceleration:
a = (13.67 m/s - 0 m/s) / 3.0 s
a ≈ 4.56 m/s²
Next, we can apply Newton's second law to find the net force:
F = 820 kg * 4.56 m/s²
F ≈ 3740.32 N
Therefore, the net force applied to the car is approximately 3740.32 N.
To find the net force applied to the car, we can use Newton's second law of motion, which states that F = ma, where F is the force, m is the mass, and a is the acceleration.
In this case, we are given the mass of the car (m = 820 kg) and the time it takes for the car to travel a certain distance (t = 3.0 s). We need to find the acceleration (a) first.
To find the acceleration, we can use the kinematic equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Since the car starts from rest (u = 0 m/s), the equation simplifies to: s = (1/2)at^2
Plugging in the values, we have: 41 m = (1/2)a(3.0 s)^2
Rearranging the equation to solve for acceleration (a), we get: a = (2s) / (t^2)
Substituting the given values, we get: a = (2 * 41 m) / (3.0 s)^2
a ≈ 9.12 m/s^2
Now that we have the acceleration, we can find the net force applied to the car by using Newton's second law of motion: F = ma
Substituting the values, we get: F = (820 kg)(9.12 m/s^2)
F ≈ 7482.4 N
Therefore, the net force applied to the car is approximately 7482.4 N.