A11kg block slides up a 30° inclined plane at a constant velocity. the coefficient of friction between the block and the plane is μk=0.2. in order for the block to slide up the incline, someone must apply a force to the block. what is the magnitude of the applied force if it points horizontally?

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane,

Friction coefficient,

Now,

Normal force that acts on the block is given by:

          (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

                      (2)

From eqn (1) and (2)

F = 94.74 N

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.