A 4.5 kg object is accelerated from rest to a
speed of 39.2 m/s in 76 s.
What average force was exerted on the ob-
ject during this period of acceleration?
Answer in units of N
a = (Vf - Vo) / t,
a = (39.2 - 0) / 76 = 0.516m/s^2.
F = ma = 4.5 * 0.516 = 2.32N.
To find the average force exerted on the object during the period of acceleration, you can use Newton's second law of motion, which states that force is equal to the mass of an object multiplied by its acceleration.
The given information in the question is:
Mass (m) = 4.5 kg
Initial velocity (u) = 0 m/s (since the object starts from rest)
Final velocity (v) = 39.2 m/s
Time (t) = 76 s
First, calculate the acceleration of the object using the formula:
Acceleration (a) = (final velocity - initial velocity) / time
a = (v - u) / t
a = (39.2 - 0) / 76
a = 0.5158 m/s² (rounded to four decimal places)
Now that we have the acceleration, we can use it to calculate the average force using Newton's second law:
Force (F) = mass × acceleration
F = m × a
F = 4.5 kg × 0.5158 m/s²
F = 2.3211 N (rounded to four decimal places)
Therefore, the average force exerted on the object during the period of acceleration is 2.3211 N.