Question

What force is needed to keeps 20kg box moving at a constant velocity when the mk is 0.3021 and the force is a pull at 20 degree to the ground

Answer 1

63.0 N

Step-by-step explanation:

We need to consider the resultant of the forces acting along the surface. We have two forces:

- The component of the pull parallel to the ground, which is given by

`F cos \theta`

where
`\theta=20^(\circ)` is the angle between the force and the ground

- The frictional force, given by

`\mu_k mg`

where
`\mu_k = 0.3021` is the coefficient of friction, m = 20 kg is the mass of the box and g=9.8 m/s^2.

The box is moving at constant velocity, this means zero acceleration, so the equation of equilibrium becomes:

`F cos \theta - \mu_k mg =0`

From which we can find the magnitude of the pull, F:

`F=(\mu_k mg)/(cos \theta)=((0.3021)(20 kg)(9.8 m/s^2))/(cos 20^(\circ))=63.0 N`