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

Answer:

63.0 N

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:

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