# A 925 N crate is being pushed across a level floor by a force of 325 N at an angle of 25 degrees above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

### Solution:

First, let’s define what a Newton (N) is: “The Newton (N) is defined as the amount of force that, when acting on a 1 kg mass, produces an acceleration of 1 m/s/s (one meter per second per second). Therefore, 1 N = 1 kg =D7 1 m/s/s.”

The force of friction opposing the 325 N force is .25 * 925 N (which is equal to 231.25 N). The force pulling the crate is 325 N * cos(25), i.e. 294.55 N. (This is the component of force acting parallel to the floor.)

Take the difference between these two and you will have the net force pulling the crate (which is 63.3 N). Using Newton’s equation F = ma, you will find the acceleration by dividing the net force by the mass of the crate (the mass is 925 =/(one standard gravity [9.80665 m/s/s], or 94.32 kg). The acceleration is therefore 63.3 N / 94.32 kg, which equals 0.671 m/s/s.

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