Dynamic Problems

Question

1. A 14.0 kg box is sliding to the left across a horizontal surface (the floor) with a speed 0f 8.00 m/s. The coefficient of kinetic friction between the box and the floor is 0.65. How much time will it take for the box to come to a complete stop?
2. What vertical force(s) is/are acting on the box? How much force is acting in the upward direction? The downward direction?
3. What horizontal force(s) is/are acting on the box? How much force is acting to the right? The left?
4. What is the net force acting on the box? (Include the magnitude and the direction of the force.)
5. What is the acceleration of the box? (Include the magnitude and the direction of the acceleration.)
6. How long will it take the box to come to a complete stop?

Answer 1

Hi Simon D.,

You've probably solved this by now, but here are some pointers:

You know the box is slowed to a stop by the force of friction.

You know the dynamic friction coefficient acts to provide a force scaled to the gravitational-normal force on the box. And that the frictional force is INDEPENDENT of the speed of the box (very important!), until the box stops.

So calculate the frictional force. Acting for __ seconds on the mass it will deliver a velocity change of -8.00 m/s, which stops the box. Calculate that ___ time.

All the rest is pretty routine, just think about where friction exerts its force.

Now here's a tougher puzzle for you. A tall thin box, if you do the same thing, tends to tip over instead of just sliding to a stop. Why? What part of the box makes it especially susceptable to tipping in this scenario? There are several areas of box to consider, but only ONE physics property which unifies them for this situation. What is that property?

-- Cheers, --Mr. d.