a 3.6 kg object is accelerated from rest to a speed of 34.4 m/s in 42 s. What average force was exerted on the object during this period of acceleration. Answer in units of N.

F=ma=mv/t

To find the average force exerted on the object, we can use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration:

Force = mass × acceleration

In this case, the mass of the object is given as 3.6 kg, and we need to find the acceleration. To do so, we can use the formula for acceleration:

acceleration = (final velocity - initial velocity) / time

Given:
Mass (m) = 3.6 kg
Final velocity (v) = 34.4 m/s
Time (t) = 42 s

Let's calculate the acceleration:

acceleration = (34.4 m/s - 0 m/s) / 42 s
acceleration = 34.4 m/s / 42 s
acceleration ≈ 0.819 m/s²

Now that we have the acceleration, we can substitute the values into Newton's second law to find the force:

Force = mass × acceleration
Force = 3.6 kg × 0.819 m/s²
Force ≈ 2.9484 kg·m/s²

The unit for force is the Newton (N). To convert the kg·m/s² to Newtons, we can use the fact that 1 kg·m/s² equals 1 N:

Force ≈ 2.9484 N

So, the average force exerted on the object during this period of acceleration is approximately 2.9484 N.