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# Magnitude of force and acceleration

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A 4.44 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=10.7N at an angle theta=22.0degrees above the horizontal, as shown. What is the magnitude of the acceleration of the block when the force is applied?

What is the magnitude of the normal force acting on the block when the force F is acting on it?

If, instead, the floor has a coefficient of kinetic friction µk = 0.04, what is the magnitude of the frictional force on the block when the block is moving?

What is the magnitude of the acceleration of the block when friction is being considered?

https://brainmass.com/physics/acceleration/magnitude-force-acceleration-146892

## SOLUTION This solution is FREE courtesy of BrainMass!

Newton's second law states that the acceleration vector is proportional to the total force vector:

Since we are dealing with vectors, this means that this relation apply to all components of the vector:

In our case the forces in the system look like:

The normal force N is the result of Newton's third law. The box pushes down on the floor, hence the floor pushes the box back perpendicular top the floor.

As you can see, we broke the pulling force into its two components.

Hence we can write two equations (one for each direction).

For the horizontal direction, the block moves with acceleration ax:

The block does not move in the vertical direction, hence the acceleration in this direction is zero.
Since we are dealing with vector components, directionality is important. It is conventional to have the vectors pointing upwards as positive.

Hence we have:

So the two equation of motion are:

For the horizontal acceleration we get:

And for the normal force:

When friction is involved, we introduce another force to the diagram:

This adds another negative component to the horizontal equation:

However, the friction force couples the two equations together, since the friction force is given by:

Where  is the friction coefficient.

Therefore our equations are now:

This leads to a horizontal acceleration of (note that the normal force has not changed):

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!