A 5.84kg box sits on a ramp that is inclined at 35.2degrees above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.27. What horizontal force is required to move the box up the incline with a constant acceleration of 4.30m/s square

N=m•g•cosα+F•sinα

ma=F•cosα-F(fr)-mgsinα =
=F•cosα - μN- mgsinα=
= F•cosα – μ(m•g•cosα+F•sinα) - mgsinα,
ma = F•cosα – μ•m•g•cosα - μ•F•sinα - mgsinα,
F• (cosα -μ•sinα) = m(a+g•sinα+μ•g•cosα),
F= m(a+g•sinα+μ•g•cosα)/ (cosα -μ•sinα).

To find the horizontal force required to move the box up the incline with a constant acceleration, we need to break down the forces acting on the box.

First, let's calculate the downward force due to gravity acting on the box:

Force of Gravity = mass x gravitational acceleration
= 5.84 kg x 9.8 m/s^2 (gravitational acceleration)
= 57.232 N

Next, let's calculate the normal force exerted by the ramp on the box. This is the force perpendicular to the inclined surface:

Normal Force = Force of Gravity x cos(theta)
= 57.232 N x cos(35.2 degrees)
= 46.693 N

The normal force is equal in magnitude and opposite in direction to the component of the weight acting perpendicular to the ramp.

Now, let's calculate the force of kinetic friction:

Force of Kinetic Friction = coefficient of kinetic friction x Normal Force
= 0.27 x 46.693 N
= 12.601 N

The force of kinetic friction opposes the motion of the box.

Finally, let's calculate the net force required to accelerate the box up the incline:

Net Force = force of kinetic friction + force required for acceleration
= 12.601 N + (mass x acceleration)
= 12.601 N + (5.84 kg x 4.30 m/s^2)
= 12.601 N + 25.112 N
= 37.713 N

The net force is the sum of the force of kinetic friction and the force required for acceleration.

Therefore, a horizontal force of 37.713 N is required to move the box up the incline with a constant acceleration of 4.30 m/s^2.