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Kryger [21]
4 years ago
10

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
Physics
1 answer:
Natali [406]4 years ago
6 0

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

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