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
average acceleration = change in speed / change it time
= (34.4 - 0) / (42 - 0 )
= .819 m/s^2
Force = mass * acceleration = 3.6 *.819 = 2.95 Newtons
To find the average force exerted on the object during this period of acceleration, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a):
F = m * a
First, let's calculate the acceleration (a):
Using the formula for acceleration (a = (vf - vi) / t), where vf is the final velocity, vi is the initial velocity (0 m/s, since the object started from rest), and t is the time taken:
a = (vf - vi) / t
a = (34.4 m/s - 0 m/s) / 42 s
a = 0.819 m/s²
Now that we have the acceleration, we can calculate the average force by multiplying the mass (m) by the acceleration (a):
F = m * a
F = 3.6 kg * 0.819 m/s²
F = 2.9484 N
Therefore, the average force exerted on the object during this period of acceleration is approximately 2.9484 N.