A 5.8 kg object is accelerated from rest to a speed of 53.3 m/s in 55 s.
What average force was exerted on the ob- ject during this period of acceleration?
Answer in units of N.
force*time=mass*changeinvelocity
force=mass(changveloicty)/time
=5.8*53.3/55 Newtons
5.62
To find the average force exerted on the object during the acceleration, you can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. The formula is:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
First, you need to calculate the acceleration (a). You can use the formula for acceleration:
a = (v - u) / t
where v is the final velocity, u is the initial velocity (which is 0 since the object starts from rest), and t is the time.
Plugging in the values, we get:
a = (53.3 m/s - 0 m/s) / 55 s
a = 53.3 m/s / 55 s
Now, calculate the value of a:
a = 0.9691 m/s^2
Next, substitute the mass (m) and acceleration (a) values into the formula for force:
F = 5.8 kg * 0.9691 m/s^2
Finally, calculate the force:
F ≈ 5.624 N
So, the average force exerted on the object during the period of acceleration is approximately 5.624 Newtons (N).