Question

The speed of a 2. 0-kg object changes from 30 m/s to 40 m/s during a 5. 0-s time interval. During this same time interval, the velocity of the object changes its direction by 90^o. What is the magnitude of the average total force acting on the object during this time interval?.

Answer 1

The magnitude of the average total force acting on the object during this time interval is equal to 20 Newton.

Given the following data:

• Mass = 2.0 kg

• Initial velocity = 30 m/s

• Final velocity = 40 m/s

• Time = 5.0 seconds.

• Angle = 90°

To calculate the magnitude of the average total force acting on the object during this time interval:

Newton's Second Law of Motion

In order to solve for the force acting on the dummy, we would apply Newton's Second Law of Motion.

Note: The acceleration of an object is equal to the rate of change in velocity with respect to time.

Mathematically, Newton's Second Law of Motion is given by this formula;

$F = \frac{M(v\;-\;u)}{t}$

Substituting the given parameters into the formula, we have;

$F_A = \frac{2(30\;-\;0)}{5}\\\\F_A = \frac{2\times 30}{5}\\\\F_A = \frac{60}{5}$

Force A = 12 Newton.

For the second force:

$F_B = \frac{2(40\;-\;0)}{5}\\\\F_B = \frac{2\times 40}{5}\\\\F_B = \frac{80}{5}$

Force B = 16 Newton.

Now, we can calculate the magnitude of the average total force acting on the object during this time interval:

$F = \sqrt{F_A^2 + F_B^2} \\\\F = \sqrt{12^2 + 16^2}\\\\F = \sqrt{144 + 256}\\\\F=\sqrt{400}$

Total force = 20 Newton.

Read more on force here: brainly.com/question/1121817