Question

A 1.5 kg rightward moving air glider decreases its speed from 1.872 meters per second to 1.798 meters per second over the length of 1 M long track. Determine the net force acting upon the air glider

Answer 1

Answer:

The magnitude of the net force is 0.204 N.

Explanation:

The net force acting upon the air glider is:

$F = ma$

Where:

m: is the mass of the air glider = 1.5 kg

a: is the acceleration

First, we need to find the acceleration. Since we have the initial and final speed we can calculate the acceleration as follows:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

$v_{f}$: is the final speed = 1.798 m/s

$v_{0}$: is the initial speed = 1.872 m/s

d: is the distance = 1 m

The acceleration is:

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(1.798 m/s)^{2} - (1.872)^{2}}{2d} = -0.136 m/s^{2}$            

The minus sing is because the air glider is desaccelerating.

Now, the magnitude of the net force is:

$|F| = ma = 1.5 kg*0.136 m/s^{2} = 0.204 N$

Therefore, the magnitude of the net force is 0.204 N.

I hope it helps you!