Question

The mass of the box on a table is 20kg. a man applied force as below the picture. the static frictional coefficient is 0.6 and the dynamic frictional coefficient is 0.5. (gravitational acceleration =10ms^{-2}

1)what is the minimum force that should be applied to move the box?
2)What is the force that should apply to move in uniform velocity?

Answer 1

Okay, let's solve this problem step-by-step:

1) To move the box initially, the applied force (F) must overcome static friction.

Static frictional force = F_static = μ_static * Weight of box

= 0.6 * 20kg * 10m/s^2

= 12N

So the minimum force to move the box is 12N.

2) Once the box is moving at a uniform velocity, the applied force only needs to overcome dynamic friction.

Dynamic frictional force = F_dynamic = μ_dynamic * Weight of box

= 0.5 * 20kg * 10m/s^2

= 10N

So to keep the box moving at a constant velocity, the applied force should be 10N.

In summary:

Minimum force to move the box (static friction) = 12N

Force to move at constant velocity (dynamic friction) = 10N

Let me know if you have any other questions!