9. A force is applied horizontally to a 20 kg box on a flat table, if the acceleration of the box is 2 m/s and the coefficient of friction between the box and the table is 0.5, what was the magnitude (number) of the
applied force?

By Newton's second law,

n + (-w) = 0

p + (-f ) = (20 kg) (2 m/s²)

where n is the magnitude of the normal force, w is the weight of the box, p is the magnitude of the applied force (p for push or pull), and f is the magnitude of the friction force.

Calculate the weight of the box:

w = (20 kg) (9.80 m/s²) = 196 N

Then

n = w = 196 N

and

f = µ n = 0.5 (196 N) = 98 N

Now solve for p :

p - 98 N = 40 N

p = 138 N

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.