How much force would be needed to push a 22 kg box across the floor with
an acceleration of 1.9 m/s^2? Assume
that there is a coefficient of friction of.
17 between the box and the floor.
Help me with thisit’s due in 10 minutes

There would be two forces acting on the box parallel to the floor, with a net force of

∑ F = p - f = m a

where p = magnitude of the push, f = mag. of friction, m = mass of the box, and a = acceleration. To find p, we first need f .

There are also only two forces acting on the box perpendicular to the floor, with net force

∑ F = n - w = 0

where n = mag. of normal force of the floor on the box and w = weight of the box. The net force is 0 because the box is only accelerating parallel to the floor.

w = m g, where g = 9.8 m/s² is the mag. of the acceleration due to gravity, so we can solve for n :

n = w = m g

n = (22 kg) (9.8 m/s²)

n = 215.6 N

The kinetic friction is proportional to the normal force by a factor of the given coefficient of friction, µ = 0.17, such that

f = µ n

f = 0.17 (215.6 N)

f = 36.652 N

Now solve for the required pushing force:

p - 36.652 N = (22 kg) (1.9 m/s²)

p ≈ 78 N

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