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?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?
To find the duration of the force applied to the car, 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 its mass and its acceleration:
F = ma
In this case, the mass of the car is given as 32 kg, and it undergoes an acceleration from 4.0 m/s to 9.0 m/s. The change in velocity (Δv) is calculated as the final velocity (v) minus the initial velocity (u):
Δv = v - u
Substituting the values given, we get:
Δv = 9.0 m/s - 4.0 m/s
Δv = 5.0 m/s
Using the equation F = ma, we can solve for the acceleration:
20 N = 32 kg x a
a = 20 N / 32 kg
a = 0.625 m/s²
Now, we have the acceleration and the change in velocity. To find the duration of the force applied, we can use the following kinematic equation:
Δv = at
Substituting the values we have:
5.0 m/s = 0.625 m/s² x t
Solving for t, we get:
t = Δv / a
t = 5.0 m/s / 0.625 m/s²
t = 8 seconds
Therefore, the force is applied for 8 seconds.
oops...got the fraction inverted
accel = force / mass = 32 / 20 = 1.6 m/s^2
(9.0 m/s - 4.0 m/s) / 1.6 m/s^2 = ? s
F = M*a.
20 = 32a,
a = 0.625 m/s^2.
V = Vo + a*t.
9 = 4 + 0.625t,
t = ?.